FS
O
PR
O
PA
G
E
4
U
N
C
O
R
R
EC
TE
D
Cubic
polynomials
4.1 Kick off with CAS
4.2 Polynomials
4.3 The remainder and factor theorems
4.4 Graphs of cubic polynomials
4.5 Equations of cubic polynomials
4.6 Cubic models and applications
4.7 Review
c04CubicPolynomials.indd 164
21/07/15 12:00 pm
4.1 Kick off with CAS
Cubic transformations
1 Using technology, sketch the following cubic functions.
2
1 3
x
e y = − x3
2
5
3
Using technology, enter y = ax into the function entry line and use a slider to
change the values of a.
Complete the following sentences.
a When sketching a cubic function, a negative sign in front of the x3 term
________ the graph of y = x3.
b When sketching a cubic function, y = ax3, for values of a < −1 and a > 1, the
graph of y = x3 becomes ___________.
c When sketching a cubic function, y = ax3, for values of −1 < a < 1, a ≠ 0, the
graph of y = x3 becomes ___________.
Using technology, sketch the following functions.
a y = x3
b y = (x + 1) 3
c y = −(x − 2) 3
d y = x3 − 1
e y = −x3 + 2
f y = 3 − x3
Using technology, enter y = (x − h) 3 into the function entry line and use a slider
to change the values of h.
Using technology, enter y = x3 + k into the function entry line and use a slider to
change the values of k.
Complete the following sentences.
a When sketching a cubic function, y = (x − h) 3, the graph of y = x3 is _________.
b When sketching a cubic function, y = x3 + k, the graph of y = x3 is _________.
Use technology and your answers to questions 1–7 to determine the equation
that could model the shape of the Bridge of Peace in Georgia. If the technology
permits, upload a photo of the bridge to make this easier.
a y = x3
d y=
7
U
N
C
O
R
R
8
TE
D
6
EC
5
PA
G
E
4
PR
O
O
3
c y = −3x3
FS
2
b y = −x3
Please refer to the Resources tab in the Prelims section of your eBookPlUs for a comprehensive
step-by-step guide on how to use your CAS technology.
c04CubicPolynomials.indd 165
21/07/15 12:00 pm
4.2 Polynomials
A polynomial is an algebraic expression in which the power of the variable is a
positive whole number. For example, 3x2 + 5x − 1 is a quadratic polynomial in
the variable x but x32 + 5x − 1 i.e. 3x−2 + 5x − 1 is not a polynomial because of the
1
3x−2 term. Similarly, !3x + 5 is a linear polynomial but 3 !x + 5 i.e. 3x 2 + 5 is
not a polynomial because the power of the variable x is not a whole number. Note
that the coefficients of x can be positive or negative integers, rational or irrational
real numbers.
Units 1 & 2
AOS 1
Topic 3
Concept 1
FS
Classification of polynomials
Polynomials
Concept summary
PR
O
O
• The degree of a polynomial is the highest power of the variable.
For example, linear polynomials have degree 1, quadratic polynomials have
degree 2 and cubic polynomials have degree 3.
• The leading term is the term containing the highest power of the variable.
• If the coefficient of the leading term is 1 then the polynomial is said to be monic.
• The constant term is the term that does not contain the variable.
PA
G
E
A polynomial of degree n has the form an xn + an−1xn−1 + . . . + a1x + a0
where n ∈ N and the coefficients an, an−1, . . . a1, a0 ∈ R. The leading term is
anxn and the constant term is a0.
Select the polynomials from the following list of algebraic expressions and for
these polynomials, state the degree, the coefficient of the leading term, the
constant term and the type of coefficients.
x4
A 5x3 + 2x2 − 3x + 4
B 5x − x3 +
C 4x5 + 2x2 + 7x−3 + 8
2
TE
D
1
EC
WORKeD
eXaMpLe
R
tHinK
1 Check the powers of the variable x
O
R
in each algebraic expression.
C
2 For polynomial A, state the degree, the
U
N
coefficient of the leading term and the
constant term.
3 Classify the coefficients of polynomial A
as elements of a subset of R.
4 For polynomial B, state the degree, the
coefficient of the leading term and the
constant term.
5 Classify the coefficients of polynomial B
as elements of a subset of R.
166
WritE
A and B are polynomials since all the powers of x
are positive integers. C is not a polynomial due to
the 7x−3 term.
Polynomial A: the leading term of 5x3 + 2x2 − 3x + 4
is 5x3.
Therefore, the degree is 3 and the coefficient of the
leading term is 5. The constant term is 4.
The coefficients in polynomial A are integers.
Therefore, A is a polynomial over Z.
x4 x4
Polynomial B: the leading term of 5x − x3 + is .
2
2
Therefore, the degree is 4 and the coefficient of the
leading term is 12 . The constant term is 0.
The coefficients in polynomial B are rational numbers.
Therefore, B is a polynomial over Q.
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 166
21/07/15 12:00 pm
polynomial notation
• The polynomial in variable x is often referred to as P(x).
• The value of the polynomial P(x) when x = a is written as P(a).
• P(a) is evaluated by substituting a in place of x in the P(x) expression.
2
a If P(x) = 5x3 + 2x2 − 3x + 4 calculate P(−1).
b If P(x) = ax2 − 2x + 7 and P(4) = 31, obtain the value of a.
tHinK
WritE
a Substitute −1 in place of x and evaluate.
a
P(x) = 5x3 + 2x2 − 3x + 4
FS
WORKeD
eXaMpLe
P(−1) = 5(−1) 3 + 2(−1) 2 − 3(−1) + 4
b 1 Find an expression for P(4) by substituting
4 in place of x, and then simplify.
PR
O
=4
O
= −5 + 2 + 3 + 4
b P(x) = ax2 − 2x + 7
P(4) = a(4) 2 − 2(4) + 7
= 16a − 1
P(4) = 31
E
2 Equate the expression for P(4) with 31.
PA
G
⇒ 16a − 1 = 31
16a = 32
a=2
TE
D
3 Solve for a.
Identity of polynomials
O
Calculate the values of a, b and c so that x(x − 7) ≡ a(x − 1) 2 + b(x − 1) + c.
U
N
tHinK
3
C
WORKeD
eXaMpLe
R
R
EC
If two polynomials are identically equal then the coefficients of like terms are equal.
Equating co-efficients means that if ax2 + bx + c ≡ 2x2 + 5x + 7 then a = 2, b = 5
and c = 7. The identically equal symbol ‘≡’ means the equality holds for all values
of x. For convenience, however, we shall replace this symbol with the equality symbol
‘=’ in working steps.
1 Expand each bracket and express both sides of
the equality in expanded polynomial form.
2 Equate the coefficients of like terms.
WritE
x(x − 7) ≡ a(x − 1) 2 + b(x − 1) + c
∴ x2 − 7x = a(x2 − 2x + 1) + bx − b + c
∴ x2 − 7x = ax2 + (−2a + b)x + (a − b + c)
Equate the coefficients.
x2 : 1 = a
....(1)
x : −7 = −2a + b ....(2)
Constant: 0 = a − b + c ....(3)
topic 4 CubIC pOLynOMIaLs
c04CubicPolynomials.indd 167
167
21/07/15 12:00 pm
3 Solve the system of simultaneous equations.
Since a = 1, substitute a = 1 into equation (2).
−7 = −2(1) + b
b = −5
Substitute a = 1 and b = −5 into equation (3).
0 = 1 − (−5) + c
4 State the answer.
FS
c = −6
∴ a = 1, b = −5, c = −6
Operations on polynomials
4
Given P(x) = 3x3 + 4x2 + 2x + m and Q(x) = 2x2 + kx − 5, find the values
of m and k for which 2P(x) − 3Q(x) = 6x3 + 2x2 + 25x − 25.
PA
G
WORKeD
eXaMpLe
E
PR
O
O
The addition, subtraction and multiplication of two or more polynomials results
in another polynomial. For example, if P(x) = x2 and Q(x) = x3 + x2 − 1, then
P(x) + Q(x) = x3 + 2x2 − 1, a polynomial of degree 3; P(x) − Q(x) = −x3 + 1, a
polynomial of degree 3; and P(x)Q(x) = x5 + x4 − x2, a polynomial of degree 5.
The operations of addition, subtraction and multiplication on polynomials have
previously been encountered.
tHinK
WritE
2P(x) − 3Q(x)
2P(x) − 3Q(x) by collecting like
terms together.
= 2(3x3 + 4x2 + 2x + m) − 3(2x2 + kx − 5)
TE
D
1 Form a polynomial expression for
EC
2 Equate the two expressions for 2P(x) − 3Q(x).
U
N
C
O
R
R
3 Calculate the values of m and k.
4 State the answer.
= 6x3 + 2x2 + (4 − 3k)x + (2m + 15)
Hence, 6x3 + 2x2 + (4 − 3k)x + (2m + 15)
= 6x3 + 2x2 + 25x − 25
Equate the coefficients of x.
4 − 3k = 25
k = −7
Equate the constant terms.
2m + 15 = −25
m = −20
Therefore m = −20, k = −7
Division of polynomials
There are several methods for performing the division of polynomials and
CAS technology computes the division readily. Here, two ‘by-hand’ methods
will be shown.
the inspection method for division
The division of one polynomial by another polynomial of equal or lesser degree can
be carried out by expressing the numerator in terms of the denominator.
168
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 168
21/07/15 12:00 pm
To divide (x + 3) by (x − 1), or to find
(x − 1) + 1 + 3 = (x − 1) + 4.
x+3
, write the numerator x + 3 as
x−1
x + 3 (x − 1) + 4
=
x−1
x−1
This expression can then be split into the sum of partial fractions as:
x + 3 (x − 1) + 4
=
x−1
x−1
The division is in the form:
4
x−1
PR
O
=1+
FS
x−1
4
+
x−1 x−1
O
=
dividend
remainder
= quotient +
divisor
divisor
PA
G
E
In the language of division, when the dividend (x + 3) is divided by the divisor
(x − 1) it gives a quotient of 1 and a remainder of 4. Note that from the division
statement xx +− 31 = 1 + x −4 1 we can write x + 3 = 1 × (x − 1) + 4.
TE
D
This is similar to the division of integers. For example, 7 divided by 2 gives a quotient
of 3 and a remainder of 1.
7
1
=3+
2
2
∴7=3×2+1
a Calculate the quotient and the remainder when (x + 7) is divided by (x + 5).
5
U
N
C
WORKeD
eXaMpLe
O
R
R
EC
This inspection process of division can be extended, with practice, to
division involving non-linear polynomials. It could be used to show that
x2 + 4x + 1 x(x − 1) + 5(x − 1) + 6
x2 + 4x + 1
6
=
=x+5+
and therefore
.
x−1
x−1
x−1
x−1
This result can be verified by checking that x2 + 4x + 1 = (x + 5)(x − 1) + 6.
b Use the inspection method to find
3x − 4
.
x+2
tHinK
WritE
a 1 Write the division of the two polynomials
a x + 7 = (x + 5) − 5 + 7
as a fraction.
2 Write the numerator in terms of the
denominator.
x+5
x+5
=
(x + 5) + 2
x+5
topic 4 CubIC pOLynOMIaLs
c04CubicPolynomials.indd 169
169
21/07/15 12:00 pm
(x + 5)
2
+
x+5
x+5
2
=1+
x+5
The quotient is 1 and the remainder is 2.
=
3 Split into partial fractions.
4 Simplify.
5 State the answer.
denominator.
b The denominator is (x + 2).
Since 3(x + 2) = 3x + 6, the numerator is
3x − 4 = 3(x + 2) − 6 − 4
∴ 3x − 4 = 3(x + 2) − 10
3x − 4 3(x + 2) − 10
=
x+2
x+2
3(x + 2)
10
=
−
(x + 2)
x+2
2 Split the given fraction into its partial
PR
O
O
fractions.
FS
b 1 Express the numerator in terms of the
10
x+2
10
3x − 4
∴
=3−
x+2
x+2
=3−
PA
G
E
3 Simplify and state the answer.
Algorithm for long division of polynomials
The inspection method of division is very efficient, particularly when the division
involves only linear polynomials. However, it is also possible to use the long-division
algorithm to divide polynomials.
TE
D
Interactivity
Division of
polynomials
int-2564
U
N
C
O
R
R
EC
The steps in the long-division algorithm are:
1.Divide the leading term of the divisor into the leading term of the dividend.
2.Multiply the divisor by this quotient.
3.Subtract the product from the dividend to form a remainder of lower
degree.
4.Repeat this process until the degree of the remainder is lower than that of
the divisor.
To illustrate this process, consider (x2 + 4x + 1) divided by (x − 1). This is written as:
x − 1 q x2 + 4x + 1
Step 1.The leading term of the divisor (x − 1) is x; the leading term of the dividend
x2
(x2 + 4x + 1) is x2. Dividing x into x2, we get = x. We write this quotient
x
x on top of the long-division symbol.
x
x − 1 q x2 + 4x + 1
170 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 170
21/07/15 12:00 pm
Step 2.The divisor (x − 1) is multiplied by the quotient x to give x(x − 1) = x2 − x.
This product is written underneath the terms of (x2 + 4x + 1); like terms are
placed in the same columns.
x
x − 1 q x2 + 4x + 1
x2 − x
Step 3.x2 − x is subtracted from (x2 + 4x + 1). This cancels out the x2 leading term
to give x2 + 4x + 1 − (x2 − x) = 5x + 1.
x
O
x2 − x
5x + 1
FS
x − 1 q x2 + 4x + 1
The division statement, so far, would be x +x −4x 1+ 1 = x + 5xx −+ 11. This is
incomplete since the remainder (5x + 1) is not of a smaller degree than the
divisor (x − 1). The steps in the algorithm must be repeated with the same
divisor (x − 1) but with (5x + 1) as the new dividend.
Continue the process.
E
PR
O
2
PA
G
Step 4. Divide the leading term of the divisor (x − 1) into the leading term of (5x + 1);
5x
this gives = 5. Write this as +5 on the top of the long-division symbol.
x
x+5
x − 1 q x2 + 4x + 1
EC
TE
D
x2 − x
5x + 1
Step 5. M
ultiply (x − 1) by 5 and write the result underneath the terms of (5x + 1).
x+5
x − 1 q x2 + 4x + 1
x − 1 q x2 + 4x + 1
x2 − x
5x + 1
5x − 5
6 ← Remainder
U
N
C
O
R
R
x2 − x
5x + 1
5x − 5
Step 6. S
ubtract (5x − 5) from (5x + 1).
x + 5 ← Quotient
The remainder is of lower degree than the divisor so no further division is possible
and we have reached the end of the process.
Thus:
x2 + 4x + 1
6
=x+5+
x−1
x−1
This method can be chosen instead of the inspection method, or if the inspection
method becomes harder to use.
Topic 4 Cubic polynomials c04CubicPolynomials.indd 171
171
21/07/15 12:00 pm
WORKeD
eXaMpLe
6
a Given P(x) = 4x3 + 6x2 − 5x + 9, use the long-division method to divide
P(x) by (x + 3) and state the quotient and the remainder.
5
b Use the long-division method to calculate the remainder when Q 3x3 + 3 x R is
divided by (5 + 3x).
tHinK
WritE
a 1 Set up the long division.
a x + 3 q 4x3 + 6x2 − 5x + 9
4x2
4x2
x + 3 q 4x3 + 6x2 − 5x + 9
PR
O
3 The second stage of the division is to multiply
O
leading term of the divisor into the leading
term of the dividend.
FS
x + 3 q 4x3 + 6x2 − 5x + 9
2 The first stage of the division is to divide the
the result of the first stage by the divisor.
Write this product placing like terms in the
same columns.
4x3 + 12x2
TE
D
the result of the second stage from the
dividend. This will yield an expression of
lower degree than the original dividend.
5 The algorithm needs to be repeated. Divide
EC
the leading term of the divisor into the
leading term of the newly formed dividend.
R
6 Multiply the result by the divisor and write
U
N
C
O
R
this product keeping like terms in the
same columns.
7 Subtract to yield an expression of lower
degree.
Note: The degree of the expression obtained
is still not less than the degree of the divisor
so the algorithm will need to be repeated
again.
172
x + 3 q 4x3 + 6x2 − 5x + 9
PA
G
4 The third stage of the division is to subtract
E
4x2
4x3 + 12x2
− 6x2 − 5x + 9
4x2 − 6x
x + 3 q 4x3 + 6x2 − 5x + 9
4x3 + 12x2
− 6x2 − 5x + 9
4x2 − 6x
x + 3 q 4x3 + 6x2 − 5x + 9
4x3 + 12x2
− 6x2 − 5x + 9
− 6x2 − 18x
4x2 − 6x
x + 3 q 4x3 + 6x2 − 5x + 9
4x3 + 12x2
− 6x2 − 5x + 9
− 6x2 − 18x
13x + 9
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 172
21/07/15 12:00 pm
4x2 − 6x + 13
8 Divide the leading term of the divisor into
the dividend obtained in the previous step.
x + 3 q 4x3 + 6x2 − 5x + 9
4x3 + 12x2
− 6x2 − 5x + 9
− 6x2 − 18x
13x + 9
4x2 − 6x + 13
this product keeping like terms in the
same columns.
x + 3 q 4x3 + 6x2 − 5x + 9
FS
9 Multiply the result by the divisor and write
4x3 + 12x2
PR
O
− 6x2 − 18x
O
− 6x2 − 5x + 9
13x + 9
13x + 39
x + 3 q 4x3 + 6x2 − 5x + 9
PA
G
10 Subtract to yield an expression of lower
E
4x2 − 6x + 13
4x3 + 12x2
− 6x2 − 5x + 9
− 6x2 − 18x
13x + 9
O
R
R
11 State the answer.
EC
TE
D
degree.
Note: The term reached is a constant so its
degree is less than that of the divisor. The
division is complete.
C
b 1 Set up the division, expressing both the
U
N
divisor and the dividend in decreasing
powers of x. This creates the columns for
like terms.
2 Divide the leading term of the divisor into
the leading term of the dividend, multiply
this result by the divisor and then subtract
this product from the dividend.
13x + 39
− 30
4x3 + 6x2 − 5x + 9
30
= 4x2 − 6x + 13 −
x+3
x+3
2
The quotient is 4x − 6x + 13 and the
remainder is −30.
5
5
b 3x3 + 3x = 3x3 + 0x2 + 3x + 0
5 + 3x = 3x + 5
3x + 5 q 3x3 + 0x2 + 53x + 0
x2
3x + 5 q 3x3 + 0x2 + 53x + 0
3x3 + 5x2
− 5x2 + 53x + 0
Topic 4 Cubic polynomials c04CubicPolynomials.indd 173
173
21/07/15 12:00 pm
x2 − 53x
3x + 5 q 3x3 + 0x2 + 53x + 0
3 Repeat the three steps of the algorithm using
the dividend created by the first application
of the algorithm.
3x3 + 5x2
− 5x2 + 53x + 0
− 5x2 −
25
x
3
10x + 0
FS
x2 − 53x + 10
3
created by the second application of the
algorithm.
3x3 + 5x2
PR
O
− 5x2 + 53x + 0
O
3x + 5 q 3x3 + 0x2 + 53x + 0
4 Repeat the algorithm using the dividend
− 5x2 − 25
x
3
PA
G
E
10x + 0
− 50
3
5
50
−
3x3 + x
3
5
10
3
= x2 − x +
+
3x + 5
3
3
3x + 5
5
10
50
= x2 − x +
−
3
3
3(3x + 5)
50
The remainder is − .
3
R
EC
TE
D
5 State the answer.
10x + 50
3
Select the polynomials from the following list of algebraic expressions and
state their degree, the coefficient of the leading term, the constant term and the
type of coefficients.
3x2 2
A 30x + 4x5 − 2x3 + 12
B
− +1
C 5.6 + 4x − 0.2x2
x
5
2 Write down a monic polynomial over R in the variable y for which the degree is
7, the coefficient of the y2 term is −!2, the constant term is 4 and the polynomial
contains four terms.
3 WE2 a If P(x) = 7x3 − 8x2 − 4x − 1 calculate P(2).
b If P(x) = 2x2 + kx + 12 and P(−3) = 0, find k.
4 If P(x) = −2x3 + 9x + m and P(1) = 2P(−1), find m.
5 WE3 Calculate the values of a, b and c so that (2x + 1)(x − 5) ≡ a(x + 1) 2 +
b(x + 1) + c.
WE1
O
1
U
N
C
PRactise
R
Exercise 4.2 Polynomials
174 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 174
21/07/15 12:00 pm
6 Express (x + 2) 3 in the form px2 (x + 1) + qx(x + 2) + r(x + 3) + t.
Given P(x) = 4x3 − px2 + 8 and Q(x) = 3x2 + qx − 7, find the values of p
and q for which P(x) + 2Q(x) = 4x3 + x2 − 8x − 6.
8 If P(x) = x4 + 3x2 − 7x + 2 and Q(x) = x3 + x + 1, expand the product P(x)Q(x)
and state its degree.
9 WE5 a Calculate the quotient and the remainder when (x − 12) is divided
by (x + 3).
4x + 7
b Use the inspection method to find
.
2x + 1
7
WE4
FS
10 Calculate the values of a, b and c if 3x2 − 6x + 5 ≡ ax(x + 2) + b(x + 2) + c
3x2 − 6x + 5
d
in the form P(x) +
where P(x) is a
x+2
x+2
polynomial and d ∈ R.
11 WE6 a Given P(x) = 2x3 − 5x2 + 8x + 6, divide P(x) by (x − 2) and state the
quotient and the remainder.
b Use the long-division method to calculate the remainder when (x3 + 10) is
divided by (1 − 2x).
12 Obtain the quotient and remainder when (x4 − 3x3 + 6x2 − 7x + 3) is divided
by (x − 1) 2.
13 Consider the following list of algebraic expressions.
x3
A 3x5 + 7x4 −
+ x2 − 8x + 12
B9 − 5x4 + 7x2 − !5x + x3
6
C "4x5 − !5x3 + !3x − 1
D2x2 (4x − 9x2)
TE
D
Consolidate
PA
G
E
PR
O
O
and hence express
5
x6 2x8
7x 4
−
+ 2−
+
F (4x2 + 3 + 7x3) 2
10
7
5
9
3x
a Select the polynomials from the list and for each of these polynomials state:
iits degree
iithe type of coefficients
iiithe leading term
ivthe constant term.
b Give a reason why each of the remaining expressions is not a polynomial.
14 Given P(x) = 2x3 + 3x2 + x − 6, evaluate the following.
a P(3)
b P(−2)
c P(1)
1
d P(0)
e P Q −2 R
f P(0.1)
2
15 If P(x) = x − 7x + 2, obtain expressions for the following.
a P(a) − P(−a)
b P(1 + h)
c P(x + h) − P(x)
16 a If P(x) = 4x3 + kx2 − 10x − 4 and P(1) = 15, obtain the value of k.
b If Q(x) = ax2 − 12x + 7 and Q(−2) = −5, obtain the value of a.
c If P(x) = x3 − 6x2 + nx + 2 and P(2) = 3P(−1), obtain the value of n.
d If Q(x) = −x2 + bx + c and Q(0) = 5 and Q(5) = 0, obtain the values of b and c.
17 a If 3x2 + 4x − 7 ≡ a(x + 1) 2 + b(x + 1) + c calculate a, b and c.
b If x3 + mx2 + nx + p ≡ (x − 2)(x + 3)(x − 4) calculate m, n and p.
c If x2 − 14x + 8 ≡ a(x − b) 2 + c calculate a, b and c and hence express
x2 − 14x + 8 in the form a(x − b) 2 + c.
d Express 4x3 + 2x2 − 7x + 1 in the form ax2 (x + 1) + bx(x + 1) + c(x + 1) + d.
U
N
C
O
R
R
EC
E
Topic 4 Cubic polynomials c04CubicPolynomials.indd 175
175
21/07/15 12:00 pm
18 aIf P(x) = 2x2 − 7x − 11 and Q(x) = 3x4 + 2x2 + 1, find, expressing the terms
in descending powers of x:
iQ(x) − P(x)
ii3P(x) + 2Q(x)
iiiP(x)Q(x)
b If P(x) is a polynomial of degree m and Q(x) is a polynomial of degree n where
m > n, state the degree of:
iP(x) + Q(x)
iiP(x) − Q(x)
iiiP(x)Q(x)
19 aDetermine the values of a, b, p and q if P(x) = x3 − 3x2 + px − 2,
R
EC
TE
D
PA
G
E
PR
O
O
FS
Q(x) = ax3 + bx2 + 3x − 2a and 2P(x) − Q(x) = 5(x3 − x2 + x + q).
b iExpress 4x4 + 12x3 + 13x2 + 6x + 1 in the form (ax2 + bx + c) 2 where a > 0.
ii
Hence state a square root of 4x4 + 12x3 + 13x2 + 6x + 1.
20 P(x) = x4 + kx2 + n2, Q(x) = x2 + mx + n and the product
P(x)Q(x) = x6 − 5x5 − 7x4 + 65x3 − 42x2 − 180x + 216.
a Calculate k, m and n.
b Obtain the linear factors of P(x)Q(x) = x6 − 5x5 − 7x4 + 65x3 − 42x2 − 180x + 216.
21 Specify the quotient and the remainder when:
a (x + 7) is divided by (x − 2)
b (8x + 5) is divided by (2x + 1)
c (x2 + 6x − 17) is divided by (x − 1)
d (2x2 − 8x + 3) is divided by (x + 2)
e (x3 + 2x2 − 3x + 5) is divided by (x − 3)
f (x3 − 8x2 + 9x − 2) is divided by (x − 1).
22 Perform the following divisions.
a (8x3 + 6x2 − 5x + 15) divided by (1 + 2x)
b (4x3 + x + 5) divided by (2x − 3)
c (x3 + 6x2 + 6x − 12) ÷ (x + 6)
d (2 + x3) ÷ (x + 1)
x4 + x3 − x2 + 2x + 5
e
x2 − 1
x(7 − 2x2)
f
(x + 2)(x − 3)
R
23a Use CAS technology to divide (4x3 − 7x2 + 5x + 2) by (2x + 3).
b State the remainder and the quotient.
U
N
C
O
Master
176 3
c Evaluate the dividend if x = −2.
3
d Evaluate the divisor if x = −2.
24a Define P(x) = 3x3 + 6x2 − 8x − 10 and Q(x) = −x3 + ax − 6.
2
3
c Give an algebraic expression for P(2n) + 24Q(n).
d Obtain the value of a so that Q(−2) = −16.
b Evaluate P(−4) + P(3) − Pa b.
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 176
21/07/15 12:01 pm
4.3 The remainder and factor theorems
The remainder obtained when dividing P(x) by the linear divisor (x − a) is of
interest because if the remainder is zero, then the divisor must be a linear factor of
the polynomial. To pursue this interest we need to be able to calculate the remainder
quickly without the need to do a lengthy division.
Units 1 & 2
the remainder theorem
Topic 3
Concept 2
PR
O
O
The remainder and
factor theorems
Concept summary
The actual division, as we know, will result in a quotient and a remainder. This is
P(x)
remainder
expressed in the division statement
= quotient +
.
x−a
x−a
Since (x − a) is linear, the remainder will be some constant term independent of x.
From the division statement it follows that:
P(x) = (x − a) × quotient + remainder
If we let x = a, this makes (x − a) equal to zero and the statement becomes:
P(a) = 0 × quotient + remainder
Therefore:
P(a) = remainder
This result is known as the remainder theorem.
FS
AOS 1
PA
G
E
Interactivity
The remainder and
factor theorems
int-2565
If a polynomial P(x) is divided by (x − a) then the
remainder is P(a).
R
Find the remainder when P(x) = x3 − 3x2 − 2x + 9 is divided by:
a x−2
C
tHinK
7
O
WORKeD
eXaMpLe
R
EC
TE
D
Note that:
• If P(x) is divided by (x + a) then the remainder would be P(−a) since replacing x
by −a would make the (x + a) term equal zero.
b
• If P(x) is divided by (ax + b) then the remainder would be Pa− b since replacing
a
b
x by − would make the (ax + b) term equal zero.
a
U
N
a 1 What value of x will make the divisor zero?
b 2x + 1.
WritE
a (x − 2) = 0 ⇒ x = 2
2 Write an expression for the remainder.
P(x) = x3 − 3x2 − 2x + 9
Remainder is P(2).
3 Evaluate to obtain the remainder.
P(2) = (2) 3 − 3(2) 2 − 2(2) + 9
=1
The remainder is 1.
b 1 Find the value of x which makes the
divisor zero.
1
b (2x + 1) = 0 ⇒ x = −
2
topic 4 CubIC pOLynOMIaLs
c04CubicPolynomials.indd 177
177
21/07/15 12:01 pm
Remainder is P Q −12 R .
2 Write an expression for the remainder and
evaluate it.
3
2
P Q −12 R = Q −12 R − 3 Q −12 R − 2 Q −12 R + 9
= −18 − 34 + 1 + 9
= 918
Remainder is 918 .
FS
the factor theorem
E
PR
O
O
We know 4 is a factor of 12 because it divides 12 exactly, leaving no remainder.
Similarly, if the division of a polynomial P(x) by (x − a) leaves no remainder, then
the divisor (x − a) must be a factor of the polynomial P(x).
P(x) = (x − a) × quotient + remainder
If the remainder is zero, then P(x) = (x − a) × quotient.
Therefore (x − a) is a factor of P(x).
This is known as the factor theorem.
PA
G
If P(x) is a polynomial and P(a) = 0 then (x − a) is a factor of P(x).
8
a Show that (x + 3) is a factor of Q(x) = 4x4 + 4x3 − 25x2 − x + 6.
b Determine the polynomial P(x) = ax3 + bx + 2 which leaves a remainder of
R
WORKeD
eXaMpLe
EC
TE
D
Conversely, if (x − a) is a factor of a polynomial P(x) then P(a) = 0.
a is a zero of the polynomial.
b
It also follows from the remainder theorem that if Pa− b = 0, then (ax + b) is a
a
−b
is a zero of the polynomial.
factor of P(x) and
a
O
R
−9 when divided by (x − 1) and is exactly divisible by (x + 2).
tHinK
C
a 1 State how the remainder can be
U
N
calculated when Q(x) is divided by
the given linear expression.
2 Evaluate the remainder.
WritE
a Q(x) = 4x4 + 4x3 − 25x2 − x + 6
When Q(x) is divided by (x + 3), the remainder
equals Q(–3).
Q(−3) = 4(−3) 4 + 4(−3) 3 − 25(−3) 2 − 1 −3 2 + 6
= 324 − 108 − 225 + 3 + 6
=0
3 It is important to explain in the answer
why the given linear expression
is a factor.
178
Since Q(−3) = 0, (x + 3) is a factor of Q(x).
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 178
21/07/15 12:01 pm
b 1 Express the given information in
terms of the remainders.
b P(x) = ax3 + bx + 2
Dividing by (x − 1) leaves a remainder of −9.
⇒ P(1) = −9
Dividing by (x + 2) leaves a remainder of 0.
⇒ P(−2) = 0
P(1) = a + b + 2
2 Set up a pair of simultaneous equations
in a and b.
a + b + 2 = −9
∴
a + b = −11..............................(1)
FS
P(−2) = −8a − 2b + 2
−8a − 2b + 2 = 0
4a + b = 1............................(2)
a + b = −11.....(1)
PR
O
3 Solve the simultaneous equations.
O
∴
4a + b = 1..........(2)
Equation (2) – equation (1):
3a = 12
PA
G
E
a=4
Substitute a = 4 into equation (1).
4 + b = −11
b = −15
TE
D
4 Write the answer.
∴ P(x) = 4x3 − 15x + 2
Factorising polynomials
U
N
C
O
R
R
EC
When factorising a cubic or higher-degree polynomial, the first step should be to check if
any of the standard methods for factorising can be used. In particular, look for a common
further on, then look to see if a grouping technique can produce either a common linear
factor or a difference of two squares. If the standard techniques do not work then the
remainder and factor theorems can be used to factorise, since the zeros of a polynomial
enable linear factors to be formed.
Cubic polynomials may have up to three zeros and therefore up to three linear factors.
For example, a cubic polynomial P(x) for which it is known that P(1) = 0, P(2) = 0
and P(−4) = 0, has 3 zeros: x = 1, x = 2 and x = −4. From these, its three linear
factors (x − 1), (x − 2) and (x + 4) are formed.
Integer zeros of a polynomial may be found through a trial-and-error process where
factors of the polynomial’s constant term are tested systematically. For the polynomial
P(x) = x3 + x2 − 10x + 8, the constant term is 8 so the possibilities to test are
1, −1, 2, −2, 4, −4, 8 and −8. This is a special case of what is known as the rational
root theorem, where if a polynomial with integer coefficients has a rational zero pq
then its constant term is divisible by p and its leading coefficient is divisible by q.
If the polynomial is not monic, then test the factors of its constant term, or test the
factors of its constant term divided by factors of its leading coefficient.
In practice, not all of the zeros need to be, nor necessarily can be, found through trial and
error. For a cubic polynomial it is sufficient to find one zero by trial and error and form its
Topic 4 Cubic polynomials c04CubicPolynomials.indd 179
179
21/07/15 12:01 pm
PA
G
E
PR
O
O
FS
corresponding linear factor using the factor theorem. Dividing this linear factor into the
cubic polynomial gives a quadratic quotient and zero remainder, so the quadratic quotient
is also a factor. The standard techniques for factorising quadratics can then be applied.
For the division step, long division could be used; however, it is more efficient to use
a division method based on equating coefficients. With practice, this can usually be
done by inspection. To illustrate, P(x) = x3 + x2 − 10x + 8 has a zero of x = 1 since
P(1) = 0. Therefore (x − 1) is a linear factor and P(x) = (x − 1)(ax2 + bx + c).
Note that the x3 term of (x − 1)(ax2 + bx + c) can only be formed by the product
of the x term in the first bracket with the x2 term in the second bracket; likewise, the
constant term of (x − 1)(ax2 + bx + c) can only be formed by the product of the
constant terms in the first and second brackets.
The coefficients of the quadratic factor are found by equating coefficients of like
terms in x3 + x2 − 10x + 8 = (x − 1)(ax2 + bx + c).
For x3: 1 = a
For constants: 8 = −c ⇒ c = −8
This gives x3 + x2 − 10x + 8 = (x − 1)(x2 + bx − 8) which can usually be written
down immediately.
For the right-hand expression (x − 1)(x2 + bx − 8), the coefficient of x2 is formed
after a little more thought. An x2 term can be formed by the product of the x term in
the first bracket with the x term in the second bracket and also by the product of the
constant term in the first bracket with the x2 term in the second bracket.
TE
D
x3 + x2 – 10x + 8 = (x – 1)(x2 + bx – 8)
= (x − 1)(x − 2)(x + 4)
a Factorise P(x) = x3 − 2x2 − 5x + 6.
b Given that (x + 1) and (5 − 2x) are factors of P(x) = −4x3 + 4x2 + 13x + 5,
C
9
U
N
WORKeD
eXaMpLe
O
R
R
EC
Equating coefficients of x2: 1 = b − 1
∴b=2
If preferred, the coefficients of x could be equated or used as check.
It follows that:
P(x) = (x − 1)(x2 + 2x − 8)
completely factorise P(x).
tHinK
WritE
a 1 The polynomial does not factorise
a P(x) = x3 − 2x2 − 5x + 6
by a grouping technique so a zero
needs to be found. The factors of the
constant term are potential zeros.
2 Use the remainder theorem to test
systematically until a zero is obtained.
Then use the factor theorem to state
the corresponding linear factor.
180
The factors of 6 are ±1, ±2, ±3 and ±6.
P(1) = 1 − 2 − 5 + 6
=0
∴ (x − 1) is a factor.
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 180
21/07/15 12:01 pm
∴ x3 − 2x2 − 5x + 6 = (x − 1)(ax2 + bx + c)
3 Express the polynomial in terms of a
product of the linear factor and a
general quadratic factor.
For the coefficient of x3 to be 1, a = 1.
For the constant term to be 6, c = −6.
∴ x3 − 2x2 − 5x + 6 = (x − 1)(x2 + bx − 6)
5 Calculate the value of b.
Equating the coefficients of x2 gives:
FS
4 State the values of a and c.
x3 – 2x2 – 5x + 6 = (x – 1)(x2 + bx – 6)
PR
O
b = −1
O
−2 = b − 1
∴ x3 − 2x2 − 5x + 6 = (x − 1)(x2 − x − 6)
Hence,
P(x) = x3 − 2x2 − 5x + 6
6 Factorise the quadratic factor so the
E
polynomial is fully factorised into its
linear factors.
PA
G
= (x − 1)(x2 − x − 6)
= (x − 1)(x − 3)(x + 2)
b 1 Multiply the two given linear factors to
TE
D
form the quadratic factor.
b P(x) = −4x3 + 4x2 + 13x + 5
2 Express the polynomial as a product
EC
of the quadratic factor and a general
linear factor.
U
N
C
O
R
R
3 Find a and b.
4 State the answer.
Since (x + 1) and (5 − 2x) are factors,
then (x + 1)(5 − 2x) = −2x2 + 3x + 5 is a
quadratic factor.
The remaining factor is linear.
∴ P(x) = (x + 1)(5 − 2x)(ax + b)
= (−2x2 + 3x + 5)(ax + b)
−4x3 + 4x2 + 13x + 5 = (−2x2 + 3x + 5)(ax + b)
Equating coefficients of x3 gives:
−4 = −2a
∴a=2
Equating constants gives:
5 = 5b
∴b=1
−4x3 + 4x2 + 13x + 5 = (−2x2 + 3x + 5)(2x + 1)
= (x + 1)(5 − 2x)(2x + 1)
∴ P(x) = (x + 1)(5 − 2x)(2x + 1)
Topic 4 Cubic polynomials c04CubicPolynomials.indd 181
181
21/07/15 12:01 pm
polynomial equations
Solve the equation 3x3 + 4x2 = 17x + 6.
tHinK
O
10
WritE
PR
O
WORKeD
eXaMpLe
FS
If a polynomial is expressed in factorised form, then the polynomial equation can be
solved using the null factor law.
(x − a)(x − b)(x − c) = 0
∴ (x − a) = 0, (x − b) = 0, (x − c) = 0
∴ x = a, x = b or x = c
x = a, x = b and x = c are called the roots or the solutions to the equation P(x) = 0.
The factor theorem may be required to express the polynomial in factorised form.
3x3 + 4x2 = 17x + 6
1 Rearrange the equation so one side is zero.
3x3 + 4x2 − 17x − 6 = 0
2 Since the polynomial does not factorise
Let P(x) = 3x3 + 4x2 − 17x − 6.
Test factors of the constant term:
P(1) ≠ 0
PA
G
E
by grouping techniques, use the remainder
theorem to find a zero and the factor theorem
to form the corresponding linear factor.
P(−1) ≠ 0
Note: It is simpler to test for integer zeros first. P(2) = 3(2) 3 + 4(2) 2 − 17(2) − 6
= 24 + 16 − 34 − 6
TE
D
=0
Therefore (x − 2) is a factor.
3 Express the polynomial as a product of the
EC
linear factor and a general quadratic factor.
R
4 Find and substitute the values of a and c.
O
R
5 Calculate b.
U
N
C
6 Completely factorise the polynomial.
7 Solve the equation.
3x3 + 4x2 − 17x − 6 = (x − 2)(ax2 + bx + c)
∴ 3x3 + 4x2 − 17x − 6 = (x − 2)(3x2 + bx + 3)
Equate the coefficients of x2:
4=b−6
b = 10
3x3 + 4x2 − 17x − 6 = (x − 2)(3x2 + 10x + 3)
= (x − 2)(3x + 1)(x + 3)
The equation 3x3 + 4x2 − 17x − 6 = 0 becomes:
(x − 2)(3x + 1)(x + 3) = 0
x − 2 = 0, 3x + 1 = 0, x + 1 = 0
1
x = 2, x = − , x = −3
3
182
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 182
21/07/15 12:01 pm
Exercise 4.3 The remainder and factor theorems
Find the remainder when P(x) = x3 + 4x2 − 3x + 5 is divided by:
a x + 2
b 2x − 1
2 If x3 − kx2 + 4x + 8 leaves a remainder of 29 when it is divided by (x − 3), find
the value of k.
3 WE8 a Show that (x − 2) is a factor of Q(x) = 4x4 + 4x3 − 25x2 − x + 6.
b Determine the polynomial P(x) = 3x3 + ax2 + bx − 2 which leaves a remainder
of −22 when divided by (x + 1) and is exactly divisible by (x − 1).
4 Given (2x + a) is a factor of 12x2 − 4x + a, obtain the value(s) of a.
5 WE9 a Factorise P(x) = x3 + 3x2 − 13x − 15.
b Given that (x + 1) and (3x + 2) are factors of P(x) = 12x3 + 41x2 + 43x + 14,
completely factorise P(x).
6 Given the zeros of the polynomial P(x) = 12x3 + 8x2 − 3x − 2 are not integers,
use the rational root theorem to calculate one zero and hence find the three linear
factors of the polynomial.
7 WE10 Solve the equation 6x3 + 13x2 = 2 − x.
8 Solve for x, 2x4 + 3x3 − 8x2 − 12x = 0.
9 Calculate the remainder without actual division when:
a x3 − 4x2 − 5x + 3 is divided by (x − 1)
b 6x3 + 7x2 + x + 2 is divided by (x + 1)
c −2x3 + 2x2 − x − 1 is divided by (x − 4)
d x3 + x2 + x − 10 is divided by (2x + 1)
e 27x3 − 9x2 − 9x + 2 is divided by (3x − 2)
f 4x4 − 5x3 + 2x2 − 7x + 8 is divided by (x − 2).
10 aWhen P(x) = x3 − 2x2 + ax + 7 is divided by (x + 2), the remainder is 11.
Find the value of a.
b If P(x) = 4 − x2 + 5x3 − bx4 is exactly divisible by (x − 1), find the value of b.
c If 2x3 + cx2 + 5x + 8 has a remainder of 6 when divided by (2x − 1), find the
value of c.
d Given that each of x3 + 3x2 − 4x + d and x4 − 9x2 − 7 have the same
remainder when divided by (x + 3), find the value of d.
11 aCalculate the values of a and b for which Q(x) = ax3 + 4x2 + bx + 1 leaves a
remainder of 39 when divided by (x − 2), given (x + 1) is a factor of Q(x).
1
b Dividing P(x) = 3x3 + mx2 + nx + 2 by either (x − 3) or (x + 3) results in the
same remainder. If that remainder is three times the remainder left when P(x) is
divided by (x − 1), determine the values of m and n.
12 aA monic polynomial of degree 3 in x has zeros of 5, 9 and –2. Express this
polynomial in:
i factorised form
ii expanded form.
b A polynomial of degree 3 has a leading term with coefficient −2 and zeros of
−4, −1 and 12. Express this polynomial in:
i factorised form
ii expanded form.
3
2
13 aGiven (x − 4) is a factor of P(x) = x − x − 10x − 8, fully factorise P(x).
b Given (x + 12) is a factor of P(x) = 3x3 + 40x2 + 49x + 12, fully factorise P(x).
c Given (5x + 1) is a factor of P(x) = 20x3 + 44x2 + 23x + 3, fully factorise P(x).
1
WE7
U
N
C
PA
G
O
R
R
EC
TE
D
Consolidate
E
PR
O
O
FS
PRactise
Topic 4 Cubic polynomials c04CubicPolynomials.indd 183
183
21/07/15 12:01 pm
d Given (4x − 3) is a factor of P(x) = −16x3 + 12x2 + 100x − 75, fully
PA
G
E
PR
O
O
FS
factorise P(x).
e Given (8x − 11) and (x − 3) are factors of P(x) = −8x3 + 59x2 − 138x + 99,
fully factorise P(x).
f Given (3x − 5) is a factor of P(x) = 9x3 − 75x2 + 175x − 125, fully
factorise P(x).
14 Factorise the following:
a x3 + 5x2 + 2x − 8
b x3 + 10x2 + 31x + 30
3
2
c 2x − 13x + 13x + 10
d −18x3 + 9x2 + 23x − 4
e x3 − 7x + 6
f x3 + x2 − 49x − 49
15 aThe polynomial 24x3 + 34x2 + x − 5 has three zeros, none of which are
integers. Calculate the three zeros and express the polynomial as the product of
its three linear factors.
5
b The polynomial P(x) = 8x3 + mx2 + 13x + 5 has a zero of .
2
iState a linear factor of the polynomial.
iiFully factorise the polynomial.
iiiCalculate the value of m.
c iFactorise the polynomials P(x) = x3 − 12x2 + 48x − 64 and Q(x) = x3 − 64.
P(x)
12x
=1− 2
iiHence, show that
.
Q(x)
x + 4x + 16
d A cubic polynomial P(x) = x3 + bx2 + cx + d has integer coefficients and
U
N
C
O
R
R
EC
TE
D
P(0) = 9. Two of its linear factors are (x − !3) and (x + !3). Calculate the
third linear factor and obtain the values of b, c and d.
16 Solve the following equations for x.
a (x + 4)(x − 3)(x + 5) = 0
b 2(x − 7)(3x + 5)(x − 9) = 0
3
2
c x − 13x + 34x + 48 = 0
d 2x3 + 7x2 = 9
e 3x2 (3x + 1) = 4(2x + 1)
f 8x4 + 158x3 − 46x2 − 120x = 0
17 aShow that (x − 2) is a factor of P(x) = x3 + 6x2 − 7x − 18 and hence fully
factorise P(x) over R.
b Show that (3x − 1) is the only real linear factor of 3x3 + 5x2 + 10x − 4.
c Show that (2x2 − 11x + 5) is a factor of 2x3 − 21x2 + 60x − 25 and hence
calculate the roots of the equation 2x3 − 21x2 + 60x − 25 = 0.
18 aIf (x2 − 4) divides P(x) = 5x3 + kx2 − 20x − 36 exactly, fully factorise P(x)
and hence obtain the value of k.
b If x = a is a solution to the equation ax2 − 5ax + 4(2a − 1) = 0, find possible
values for a.
c The polynomials P(x) = x3 + ax2 + bx − 3 and Q(x) = x3 + bx2 + 3ax − 9
have a common factor of (x + a). Calculate a and b and fully factorise each
polynomial.
d (x + a) 2 is a repeated linear factor of the polynomial P(x) = x3 + px2 + 15x + a2.
Show there are two possible polynomials satisfying this information and,
for each, calculate the values of x which give the roots of the equation
x3 + px2 + 15x + a2 = 0.
19 Specify the remainder when (9 + 19x − 2x2 − 7x3) is divided by (x − !2 + 1).
20 Solve the equation 10x3 − 5x2 + 21x + 12 = 0 expressing the values of x to 4
decimal places.
Master
184 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 184
21/07/15 12:01 pm
4.4
Units 1 & 2
AOS 1
Topic 3
Graphs of cubic polynomials
The graph of the general cubic polynomial has an equation of the form
y = ax3 + bx2 + cx + d, where a, b, c and d are real constants and a ≠ 0. Since
a cubic polynomial may have up to three linear factors, its graph may have up
to three x-intercepts. The shape of its graph is affected by the number of
x-intercepts.
The graph of y = x3 and transformations
Concept 3
The graph of the simplest cubic polynomial has
y
y = x3
3
the equation y = x .
10
The ‘maxi–min’ point at the origin is sometimes
referred to as a ‘saddle point’. Formally, it
is called a stationary point of inflection (or
x
–3 –2 –1 0
1
2
3
inflexion as a variation of spelling). It is a key
–10
feature of this cubic graph.
Key features of the graph of y = x3:
• (0, 0) is a stationary point of inflection.
• The shape of the graph changes from concave down to concave up at the
stationary point of inflection.
• There is only one x-intercept.
• As the values of x become very large positive, the behaviour of the graph
shows its y-values become increasingly large positive also. This means that
as x → ∞, y → ∞. This is read as ‘as x approaches infinity, y approaches
infinity’.
• As the values of x become very large negative, the behaviour of the graph
shows its y-values become increasingly large negative. This means that as
x → −∞, y → −∞.
• The graph starts from below the x-axis and increases as x increases.
Once the basic shape is known, the graph can be dilated, reflected and translated
in much the same way as the parabola y = x2.
PR
O
O
FS
Graphs of cubic
polynomials
Concept summary
R
EC
TE
D
PA
G
E
Interactivity
Graph plotten:
Cubic polynomial
int-2566
U
N
C
O
R
Dilation
The graph of y = 4x3 will be
narrower than the graph of y = x3
due to the dilation factor of 4 from
the x-axis. Similarly, the graph of
y = 14x3 will be wider than the graph
of y = x3 due to the dilation factor of
1
from the x-axis.
4
y
6
y = 4x3
y = x3
1
y = – x3
4
4
2
–3
–2
–1 0
1
2
3
x
–2
–4
–6
Topic 4 Cubic polynomials c04CubicPolynomials.indd 185
185
21/07/15 12:01 pm
Reflection
The graph of y = −x3 is the reflection of the
graph of y = x3 in the x-axis.
For the graph of y = −x3 note that:
• as x → ∞, y → −∞ and as
x → −∞, y → ∞
• the graph starts from above the x-axis and
decreases as x increases
• at (0, 0), the stationary point of inflection,
the graph changes from concave up to
concave down.
y = –x3
x
8
PR
O
y = (x + 4)3
O
FS
0
y = x3 + 4
(0, 4)
E
4
PA
G
translation
The graph of y = x3 + 4 is obtained
when the graph of y = x3 is
translated vertically upwards by
4 units. The stationary point of
inflection is at the point (0, 4).
The graph of y = (x + 4) 3 is
obtained when the graph of y = x3 is
translated horizontally 4 units to the
left. The stationary point of inflection
is at the point (−4, 0).
The transformations from the basic
parabola y = x2 are recognisable
from the equation y = a(x − h) 2 + k,
and the equation of the graph of
y = x3 can be transformed to a
similar form.
y
(–4, 0)
R
EC
TE
D
0
U
N
C
O
R
The key features of the graph of y = a(x − h) 3 + k are:
•stationary point of inflection at (h, k)
•change of concavity at the stationary point of inflection
•if a > 0, the graph starts below the x-axis and increases, like y = x3
•if a < 0, the graph starts above the x-axis and decreases, like y = −x3
•the one x-intercept is found by solving a(x − h) 3 + k = 0
•the y-intercept is found by substituting x = 0.
WORKeD
eXaMpLe
11
Sketch
1
2
b y = 6 − (x − 2) 3
a y = (x + 1) 3 + 8
tHinK
WritE
a 1 State the point of inflection.
a y = (x + 1) 3 + 8
Point of inflection is (−1, 8).
186
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 186
21/07/15 12:01 pm
y-intercept: let x = 0
y = (1) 3 + 8
2 Calculate the y-intercept.
=9
⇒ (0, 9)
x-intercept: let y = 0
3 Calculate the x-intercept.
(x + 1) 3 + 8 = 0
FS
(x + 1) 3 = −8
Take the cube root of both sides:
3
x+1="
−8
x = −3
⇒ (−3, 0)
y
PA
G
points and ensure the graph
changes concavity at the point of
inflection.
The coefficient of x3 is positive so the graph starts
below the x-axis and increases.
E
4 Sketch the graph. Label the key
PR
O
O
x + 1 = −2
(–1, 8)
y = (x + 1)3 + 8
(0, 9)
(–3, 0)
TE
D
0
b 1 Rearrange the equation to the
R
EC
y = a(x − h) 3 + k form and state
the point of inflection.
U
N
C
O
R
2 Calculate the y-intercept.
3 Calculate the x-intercept.
Note: A decimal approximation
helps locate the point.
x
b y = 6 − 1 (x − 2) 3
2
= −12 (x − 2) 3 + 6
Point of inflection: (2, 6)
y-intercept: let x = 0
1
y = − (−2) 3 + 6
2
= 10
⇒ (0, 10)
x-intercept: let y = 0
−12 (x − 2) 3 + 6 = 0
1
(x
2
− 2) 3 = 6
(x − 2) 3 = 12
3
x−2="
12
3
x=2+"
12
3
⇒ (2 + "
12, 0) ≈ (4.3, 0)
Topic 4 Cubic polynomials c04CubicPolynomials.indd 187
187
21/07/15 12:01 pm
a < 0 so the graph starts above the x-axis and decreases.
4 Sketch the graph showing all
key features.
y
1
y = 6– – (x – 2)3
2
(0, 10)
(2, 6)
5
(2 + 12, 0)
x
FS
0
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
Cubic graphs with one x-intercept but no stationary
point of inflection
y
There are cubic graphs
which have one x-intercept
y = (x + 1) (x2 + x + 3)
y = (x – 4) (x2 + x + 3)
but no stationary point of
inflection. The equations of
such cubic graphs cannot
(0, 3)
(–1, 0)
be expressed in the form
x
0
(4, 0)
y = a(x − h) 3 + k. Their
equations can be expressed
as the product of a linear
(0, –12)
factor and a quadratic
factor which is irreducible,
meaning the quadratic has
no real factors.
Technology is often required to sketch such graphs. Two examples,
y = (x + 1)(x2 + x + 3) and y = (x − 4)(x2 + x + 3), are shown in the diagram. Each
has a linear factor and the discriminant of the quadratic factor x2 + x + 3 is negative;
this means it cannot be further factorised over R.
Both graphs maintain the long-term behavior exhibited by all cubics with a positive
leading-term coefficient; that is, as x → ∞, y → ∞ and as x → −∞, y → −∞.
Every cubic polynomial must have at least one linear factor in order to maintain this
long-term behaviour.
188 Cubic graphs with three x-intercepts
For the graph of a cubic polynomial to have three x-intercepts, the polynomial must
have three distinct linear factors. This is the case when the cubic polynomial expressed
as the product of a linear factor and a quadratic factor is such that the quadratic factor
has two distinct linear factors.
y = (x – a) (x – b) (x – c)
This means that the graph of a
monic cubic with an equation of
the form y = (x − a)(x − b)(x − c)
a
x
c
b
where a, b, c ∈ R and a < b < c will
have the shape of the graph shown.
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 188
21/07/15 12:01 pm
WORKeD
eXaMpLe
12
PR
O
O
FS
If the graph is reflected in the
y = – (x – a) (x – b) (x – c)
x-axis, its equation is of the form
y = −(x − a)(x − b)(x − c) and the
shape of its graph satisfies the
x
c
a
b
long-term behaviour that as
x → ±∞, y → ∓∞.
It is important to note the graph
is not a quadratic so the maximum and minimum turning points do not lie halfway
between the x-intercepts. In a later chapter we will learn how to locate these points
without using technology.
To sketch the graph, it is usually sufficient to identify the x- and y-intercepts and
to ensure the shape of the graph satisfies the long-term behaviour requirement
determined by the sign of the leading term.
Sketch the following without attempting to locate turning points.
a y = (x − 1)(x − 3)(x + 5)
b y = (x + 1)(2x − 5)(6 − x)
WritE
a 1 Calculate the x-intercepts.
a y = (x − 1)(x − 3)(x + 5)
E
tHinK
PA
G
x-intercepts: let y = 0
(x − 1)(x − 3)(x + 5) = 0
x = 1, x = 3, x = −5
⇒ (−5, 0), (1, 0), (3, 0) are the x-intercepts.
3 Determine the shape of
R
the graph.
y-intercept: let x = 0
y = (−1)(−3)(5)
= 15
⇒ (0, 15) is the y-intercept.
TE
D
EC
2 Calculate the y-intercept.
Multiplying together the terms in x from each bracket
gives x3, so its coefficient is positive.
The shape is of a positive cubic.
y
U
N
C
O
R
4 Sketch the graph.
b 1 Calculate the x-intercepts.
y = (x – 1) (x – 3) (x + 5)
(0, 15)
(–5, 0)
0
(3, 0)
(1, 0)
x
b y = (x + 1)(2x − 5)(6 − x)
x-intercepts: let y = 0
(x + 1)(2x − 5)(6 − x) = 0
x + 1 = 0, 2x − 5 = 0, 6 − x = 0
x = −1, x = 2.5, x = 6
⇒ (−1, 0), (2.5, 0), (6, 0) are the x-intercepts.
topic 4 CubIC pOLynOMIaLs
c04CubicPolynomials.indd 189
189
21/07/15 12:01 pm
2 Calculate the y-intercept.
y-intercept: let x = 0
y = (1)(−5)(6)
= −30
⇒ (0, −30) is the y-intercept.
3 Determine the shape of
Multiplying the terms in x from each bracket gives
(x) × (2x) × (−x) = −2x3 so the shape is of a negative cubic.
the graph.
4 Sketch the graph.
(6, 0)
x
(2.5, 0)
(–1, 0)
PR
O
Cubic graphs with two x-intercepts
O
0
(0, –30)
FS
y = (x + 1) (2x – 5) (6 – x)
PA
G
E
If a cubic has two x-intercepts one at x = a and one at x = b then in order to satisfy
the long-term behaviour required of any cubic, the graph either touches the x-axis at
x = a and turns, or it touches the x-axis at x = b and turns. One of the x-intercepts
must be a turning point.
a
b
y = (x – a) (x – b)2
x
R
R
EC
TE
D
y = (x – a)2 (x – b)
U
N
C
O
Thinking of the cubic polynomial as the product of a linear and a quadratic factor, for its
graph to have two instead of three x-intercepts, the quadratic factor must have two identical
factors. Either the factors of the cubic are (x − a)(x − a)(x − b) = (x − a) 2 (x − b) or
the factors are (x − a)(x − b)(x − b) = (x − a)(x − b) 2. The repeated factor identifies
the x-intercept which is the turning point. The repeated factor is said to be of multiplicity 2
and the single factor of multiplicity 1.
The graph of a cubic polynomial with equation of the form y = (x − a) 2 (x − b) has
a turning point on the x-axis at (a, 0) and a second x-intercept at (b, 0). The graph is
said to touch the x-axis at x = a and cut it at x = b.
Although the turning point on the x-axis must be identified when sketching the graph,
there will be a second turning point that cannot yet be located without technology.
Note that a cubic graph whose equation has a repeated factor of multiplicity 3,
such as y = (x − h) 3, would have only one x-intercept as this is a special case of
y = a(x − h) 3 + k with k = 0. The graph would cut the x-axis at its stationary point
of inflection (h, 0).
190 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 190
21/07/15 12:01 pm
WORKeD
eXaMpLe
13
Sketch the graphs of:
1
a y = (x − 2) 2 (x + 5)
4
b y = −2(x + 1)(x + 4) 2
tHinK
WritE
a 1 Calculate the x-intercepts and
a y = 1 (x − 2) 2 (x + 5)
4
interpret the multiplicity of
each factor.
x-intercepts: let y = 0
1
(x
4
− 2) 2 (x + 5) = 0
FS
∴ x = 2 (touch), x = −5 (cut)
x-intercept at (−5, 0) and turning-point x-intercept at (2, 0)
O
y-intercept: let x = 0
2 Calculate the y-intercept.
y = 14 (−2) 2 (5)
PR
O
=5
⇒ (0, 5)
y
1 (x – 2)2 (x + 5)
y = ––
4
PA
G
E
3 Sketch the graph.
(0, 5)
(–5, 0)
EC
R
R
O
U
N
C
2 Calculate the y-intercept.
x
(2, 0)
TE
D
b 1 Calculate the x-intercepts and
interpret the multiplicity of
each factor.
0
b y = −2(x + 1)(x + 4) 2
x-intercepts: let y = 0
−2(x + 1)(x + 4) 2 = 0
(x + 1)(x + 4) 2 = 0
∴ x = −1 (cut), x = −4 (touch)
x-intercept at (−1, 0) and turning-point x-intercept at
(−4, 0)
y-intercept: let x = 0
y = −2(1)(4) 2
= −32
y-intercept at (0, −32)
y
3 Sketch the graph.
y = –2(x + 1) (x + 4)2
(–4, 0)
(–1, 0)
0
x
(0, –32)
topic 4 CubIC pOLynOMIaLs
c04CubicPolynomials.indd 191
191
21/07/15 12:01 pm
Cubic graphs in the general form y = ax3 + bx2 + cx + d
14
Sketch the graph of y = x3 − 3x − 2, without attempting to obtain any
turning points that do not lie on the coordinate axes.
tHinK
WritE
TE
D
WORKeD
eXaMpLe
PA
G
E
PR
O
O
FS
If the cubic polynomial with equation y = ax3 + bx2 + cx + d can be factorised, then
the shape of its graph and its key features can be determined. Standard factorisation
techniques such as grouping terms together may be sufficient, or the factor theorem
may be required in order to obtain the factors.
The sign of a, the coefficient of x3, determines the long-term behaviour the graph
exhibits. For a > 0 as x → ±∞, y → ±∞; for a < 0 as x → ±∞, y → ∓∞.
The value of d determines the y-intercept.
The factors determine the x-intercepts and the multiplicity of each factor will
determine how the graph intersects the x-axis.
Every cubic graph must have at least one x-intercept and hence the polynomial must
have at least one linear factor. Considering a cubic as that’s the product of a linear and
a quadratic factor, it is the quadratic factor which determines whether there is more
than one x-intercept.
Graphs which have only one x-intercept may be of the form y = a(x − h) 3 + k where
the stationary point of inflection is a major feature. Recognition of this equation from
its expanded form would require the expansion of a perfect cube to be recognised,
since a(x3 − 3x2h + 3xh2 − h3) + k = a(x − h) 3 + k. However, as previously noted,
not all graphs with only one x-intercept have a stationary point of inflection.
1 Obtain the y-intercept first since it is simpler
to obtain from the expanded form.
EC
2 Factorisation will be needed in order to obtain
the x-intercepts.
R
3 The polynomial does not factorise by
U
N
C
O
R
grouping so the factor theorem needs
to be used.
y = x3 − 3x − 2
y-intercept: (0, −2)
x-intercepts: let y = 0
x3 − 3x − 2 = 0
Let P(x) = x3 − 3x − 2
P(1) ≠ 0
P(−1) = −1 + 3 − 2 = 0
∴ (x + 1) is a factor
x3 − 3x − 2 = (x + 1)(x2 + bx − 2)
= (x + 1)(x2 − x − 2)
= (x + 1)(x − 2)(x + 1)
= (x + 1) 2 (x − 2)
∴ x3 − 3x − 2 = 0
⇒ (x + 1) 2 (x − 2) = 0
∴ x = −1, 2
4 What is the nature of these x-intercepts?
192
y = P(x) = (x + 1) 2 (x − 2)
x = −1 (touch) and x = 2 (cut)
Turning point at (−1, 0)
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 192
21/07/15 12:01 pm
5 Sketch the graph.
y = x3 –3x – 2
(–1, 0)
0
(2, 0)
Exercise 4.4 Graphs of cubic polynomials
WE11
Sketch the graphs of these polynomials.
PR
O
1
1
(x + 6) 3
36
2 State the coordinates of the point of inflection and sketch the graph of the
following.
3
x
a y = a − 3b
b y = 2x3 − 2
2
a y = (x − 1) 3 − 8
b y = 1 −
3
Sketch the following, without attempting to locate turning points.
a y = (x + 1)(x + 6)(x − 4)
b y = (x − 4)(2x + 1)(6 − x)
WE12
TE
D
4 Sketch y = 3x(x2 − 4).
Sketch the graphs of these polynomials.
1
a y = (x − 3) 2 (x + 6)
b y = −2(x − 1)(x + 2) 2
9
WE13
EC
5
PA
G
E
PRactise
O
FS
(0, –2)
6 Sketch y = 0.1x(10 − x) 2 and hence shade the region for which
Sketch the graph of y = x3 − 3x2 − 10x + 24 without attempting to obtain
any turning points that do not lie on the coordinate axes.
WE14
R
7
R
y ≤ 0.1x(10 − x) 2.
U
N
C
O
8 a
Sketch the graph of y = −x3 + 3x2 + 10x − 30 without attempting to obtain any
Consolidate
turning points that do not lie on the coordinate axes.
b Determine the coordinates of the stationary point of inflection of the graph with
equation y = x3 + 3x2 + 3x + 2 and sketch the graph.
9 a
Sketch the graphs of y = x3, y = 3x3, y = x3 + 3 and y = (x + 3) 3 on the one
set of axes.
b Sketch the graphs of y = −x3, y = −3x3, y = −x3 + 3 and y = −(x + 3) 3 on the
one set of axes.
10 Sketch the graphs of the following, identifying all key points.
a y = (x + 4) 3 − 27
b y = 2(x − 1) 3 + 10
c y = 27 + 2(x − 3) 3
d y = 16 − 2(x + 2) 3
3
4
e y = − (3x + 4) 3
f y = 9 +
x3
3
Topic 4 Cubic polynomials c04CubicPolynomials.indd 193
193
21/07/15 12:01 pm
11 Sketch the graphs of the following, without attempting to locate any turning points
that do not lie on the coordinate axes.
a y = (x − 2)(x + 1)(x + 4)
b y = −0.5x(x + 8)(x − 5)
c y = (x + 3)(x − 1)(4 − x)
d y = (2 − x)(6 − x)(4 + x)
e y = 0.1(2x − 7)(x − 10)(4x + 1)
f y = 2a − 1b a
1
4
x
2
5
3x
+ 2b ax − b
4
8
12 Sketch the graphs of the following, without attempting to locate any turning points
c y = (x + 3) 2 (4 − x)
d y = 4 (2 − x) 2 (x − 12)
f y = −0.25x2 (2 − 5x)
1
e y = 3x(2x + 3) 2
FS
that do not lie on the coordinate axes.
a y = −(x + 4) 2 (x − 2)
b y = 2(x + 3)(x − 3) 2
O
13 Sketch the graphs of the following, showing any intercepts with the coordinate
PR
O
axes and any stationary point of inflection.
a y = (x + 3) 3
b y = (x + 3) 2 (2x − 1)
c y = (x + 3)(2x − 1)(5 − x)
d 2(y − 1) = (1 − 2x) 3
1
f y = −2 (2 − 3x)(3x + 2)(3x − 2)
e 4y = x(4x − 1) 2
PA
G
equations given by:
a y = 9x2 − 2x3
c y = 9x2 − 3x3 + x − 3
e y = 9x3 + 27x2 + 27x + 9
E
14 Factorise, if possible, and then sketch the graphs of the cubic polynomials with
b y = 9x3 − 4x
d y = 9x(x2 + 4x + 3)
f y = −9x3 − 9x2 + 9x + 9
TE
D
15 Sketch, without attempting to locate any turning points that do not lie on the
coordinate axes.
a y = 2x3 − 3x2 − 17x − 12
c y = x3 − 17x + 4
b y = 6 − 55x + 57x2 − 8x3
e y = −5x3 − 7x2 + 10x + 14
f y = − x3 + 14x − 24
EC
d y = 6x3 − 13x2 − 59x − 18
1
2
R
16 Consider P(x) = 30x3 + kx2 + 1.
U
N
C
O
R
a Given (3x − 1) is a factor, find the value of k.
b Hence express P(x) as the product of its linear factors.
c State the values of x for which P(x) = 0.
d Sketch the graph of y = P(x).
e Does the point (−1, −40) lie on the graph of y = P(x)? Justify your answer.
f On your graph shade the region for which y ≥ P(x).
1
17 a Express −2x3 + 6x2 − 24x + 38 in the form a(x − b) 3 + c.
1
b Hence sketch the graph of y = −2x3 + 6x2 − 24x + 38.
18 Consider y = x3 − 5x2 + 11x − 7.
a Show that the graph of y = x3 − 5x2 + 11x − 7 has only one x-intercept.
b Show that y = x3 − 5x2 + 11x − 7 cannot be expressed in the form
y = a(x − b) 3 + c.
c Describe the behaviour of the graph as x → ∞.
d Given the graph of y = x3 − 5x2 + 11x − 7 has no turning points, draw a sketch
of the graph.
194 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 194
21/07/15 12:02 pm
MAstEr
19 a Sketch, locating turning points, the graph of y = x3 + 4x2 − 44x − 96.
b Show that the turning points are not placed symmetrically in the interval
between the adjoining x-intercepts.
20 Sketch, locating intercepts with the coordinate axes and any turning points.
Express values to 1 decimal place where appropriate.
a y = 10x3 − 20x2 − 10x − 19
b y = −x3 + 5x2 − 11x + 7
c y = 9x3 − 70x2 + 25x + 500
PR
O
Units 1 & 2
AOS 1
Topic 3
As a guide:
• If there is a stationary point of inflection given, use the y = a(x − h) 3 + k
form.
• If the x-intercepts are given, use the y = a(x − x1)(x − x2)(x − x3) form,
or the repeated factor form y = a(x − x1) 2 (x − x2) if there is a turning
point at one of the x-intercepts.
• If four points on the graph are given, use the y = ax3 + bx2 + cx + d
form.
Concept 4
PA
G
E
Equations of cubic
polynomials
Concept summary
Determine the equation for each of the following graphs.
EC
15
TE
D
Interactivity
Graph plotten:
1, 2, 3 intercepts
int-2567
WORKeD
eXaMpLe
FS
The equation y = ax3 + bx2 + cx + d contains four unknown coefficients that need to
be specified, so four pieces of information are required to determine the equation of a
cubic graph, unless the equation is written in turning point form when 3 pieces of
information are required.
O
4.5
Equations of cubic polynomials
a The graph of a cubic polynomial which has a stationary point of inflection
y
c
U
N
C
O
R
b
R
at the point (−7, 4) and an x-intercept at (1, 0).
(–1, 0)
(4, 0)
0
(3, 36)
x
(–3, 0)
(0, –5)
(0, 0) (2, 0)
x
0
tHinK
WritE
a 1 Consider the given information and choose
a Stationary point of inflection is given.
the form of the equation to be used.
Let y = a(x − h) 3 + k
Point of inflection is (−7, 4).
∴ y = a(x + 7) 3 + 4
topic 4 CubIC pOLynOMIaLs
c04CubicPolynomials.indd 195
195
21/07/15 12:02 pm
2 Calculate the value of a.
Substitute the given x-intercept point (1, 0).
0 = a(8) 3 + 4
(8) 3a = −4
−4
a=
8 × 64
1
a=−
128
Note: The coordinates of the given
points show the y-values decrease as the
x-values increase, so a negative value for a
is expected.
1
(x + 7) 3 + 4.
128
b Two x-intercepts are given.
One shows a turning point at x = 4 and the
other a cut at x = −1.
Let the equation be y = a(x + 1)(x − 4) 2.
The equation is y = −
3 Write the equation of the graph.
FS
b 1 Consider the given information and choose
O
the form of the equation to be used.
PR
O
Substitute the given y-intercept point (0, −5).
−5 = a(1)(−4) 2
2 Calculate the value of a.
a=−
5
16
E
−5 = a(16)
5
(x + 1)(x − 4) 2.
The equation is y = −16
PA
G
3 Write the equation of the graph.
c 1 Consider the given information and choose
c Three x-intercepts are given.
the form of the equation to be used.
Let the equation be
y = a(x + 3)(x − 0)(x − 2)
TE
D
∴ y = ax(x + 3)(x − 2)
Substitute the given point (3, 36).
36 = a(3)(6)(1)
36 = 18a
EC
2 Calculate the value of a.
a=2
The equation is y = 2x(x + 3)(x − 2).
R
3 Write the equation of the graph.
R
Cubic inequations
U
N
Interactivity
Cubic inequations
int-2568
C
O
The sign diagram for a cubic polynomial can be deduced from the shape of its graph
and its x-intercepts. The values of the zeros of the polynomial are those of the
x-intercepts. For a cubic polynomial with a positive coefficient of x3, the sign diagram
starts from below the x-axis. For examples below, assume a < b < c.
One zero (x − a) 3 or (x − a) × irreducible quadratic factor
+
−
Two zeros (x − a) 2 (x − b)
x
a
+
−
a
x
b
Three zeros (x − a)(x − b)(x − c)
+
−
196 a
b
c
x
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 196
21/07/15 12:02 pm
At a zero of multiplicity 2, the sign diagram touches the x-axis. For a zero of odd
multiplicity, either multiplicity 1 or multiplicity 3, the sign diagram cuts the x-axis.
This ‘cut and touch’ nature applies if the coefficient of x3 is negative; however, the
sign diagram would start from above the x-axis in that case.
O
FS
To solve a cubic inequation:
• Rearrange the terms in the inequation, if necessary, so that one side of the
inequation is 0.
• Factorise the cubic expression and calculate its zeros.
• Draw the sign diagram, or the graph, of the cubic.
• Read from the sign diagram the set of values of x which satisfy the
inequation.
16
Solve the inequations.
a (x + 2)(x − 1)(x − 5) > 0
c (x − 2) 3 − 1 > 0
tHinK
PA
G
WORKeD
eXaMpLe
E
PR
O
An exception applies to inequations of forms such as a(x − h) 3 + k > 0. These
inequations are solved in a similar way to solving linear inequations without the need
for a sign diagram or a graph. Note the similarity between the sign diagram of the
cubic polynomial with one zero and the sign diagram of a linear polynomial.
b {x : 4x2 ≤ x3}
WritE
TE
D
a 1 Read the zeros from each factor.
a (x + 2)(x − 1)(x − 5) > 0
Zeros: x = −2, x = 1, x = 5
2 Consider the leading-term coefficient to draw the
+
−
EC
sign diagram. This is a positive cubic.
R
4 Show the solution on a graph.
C
O
R
40
30
20
(−2, 0) 10
–4 –3 –2 –1 0
–10
–20
–30
–40
U
N
2 Factorise to calculate the zeros.
1
5
x
−2 < x < 1 or x > 5
3 State the intervals in which the required solutions lie.
b 1 Rearrange so one side is 0.
−2
b
(1, 0) (5, 0)
1 2 3 4 5 6 7
4x2 ≤ x3
4x2 − x3 ≤ 0
∴ x2 (4 − x) ≤ 0
Let x2 (4 − x) = 0
x2 = 0 or 4 − x = 0
x = 0 or x = 4
Zeros: x = 0 (multiplicity 2), x = 4
topic 4 CubIC pOLynOMIaLs
c04CubicPolynomials.indd 197
197
21/07/15 12:02 pm
3 Consider the leading-term coefficient to draw the sign
diagram. 4x2 − x3 is a negative cubic.
+
−
0
x
4
5x : x ≥ 46 ∪ 506
4 State the answer from the sign diagram, using
set notation.
y
40
30
20
(0, 0)
10
x
FS
1 2 3 4 5
PR
O
–3 –2 –1 0
–10
–20
–30
–40
O
5 Show the solution on a graph.
c (x − 2) 3 − 1 > 0
c 1 Solve for x.
E
(x − 2) 3 > 1
3
(x − 2) > "
1
x−2>1
x>3
TE
D
PA
G
2 Show the solution on a graph.
y
15
10
5
–2 –1 0
–5
–10
–15
(3, 0)
1 2 3 4 5
x
EC
Intersections of cubic graphs with linear and quadratic
graphs
17
U
N
WORKeD
eXaMpLe
C
O
R
R
If P(x) is a cubic polynomial and Q(x) is either a linear or a quadratic polynomial, then
at the intersections of the graphs of y = P(x) and y = Q(x), P(x) = Q(x). Hence the
x-coordinates of the points of intersection are the roots of the equation P(x) − Q(x) = 0.
This is a cubic equation since P(x) − Q(x) is a polynomial of degree 3.
Sketch the graphs of y = x(x − 1)(x + 1) and y = x and calculate the
coordinates of the points of intersection. Hence state the values of x for which
x > x(x − 1)(x + 1).
tHinK
1 Sketch the graphs.
198
WritE
y = x(x − 1)(x + 1)
This is a positive cubic.
x-intercepts: let y = 0
x(x − 1)(x + 1) = 0
x = 0, x = ±1
(−1, 0), (0, 0), (1, 0) are the three x-intercepts.
y-intercept is (0, 0).
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 198
21/07/15 12:02 pm
Line: y = x passes through (0, 0) and (1, 1).
y
y = x(x – 1) (x + 1)
y=x
( 2, 2)
x
0
(0, 0)
FS
(– 2, – 2)
At intersection:
x(x − 1)(x + 1) = x
x(x2 − 1) − x = 0
x3 − 2x = 0
x(x2 − 2) = 0
x = 0, x2 = 2
x = 0, x = ±!2
O
2 Calculate the coordinates of the points of
E
PR
O
intersections.
PA
G
Substituting these x-values in the equation of
the line y = x, the points of intersection are
(0, 0), 1 !2, !2 2 , 1 −!2, −!2 2 .
3 Show the regions of the graph where
EC
TE
D
x > x(x − 1)(x + 1)
(0, 0)
y
6
4
2
( 2, 2)
x
–3.5–3–2.5–2–1.5–1–0.50 0.5 1 1.5 2 2.5 3 3.5
–2
–4
–6
R
(– 2, – 2)
x > x(x − 1)(x + 1) for
U x: x < "2 V ∪ U x: 0 < x −"2 V
4 Use the diagram to state the intervals for
O
R
which the given inequation holds.
U
N
C
Exercise 4.5 Equations of cubic polynomials
PRactise
1
Determine the equation of each of the following graphs.
a The graph of a cubic polynomial which has a stationary point of inflection at
the point (3, −7) and an x-intercept at (10, 0).
WE15
b c
y
y
(0, 12)
(−5, 0)
(4, 0)
0
(2, −7)
x
(3, 0)
(–2, 0)
x
0
Topic 4 Cubic polynomials c04CubicPolynomials.indd 199
199
21/07/15 12:02 pm
2 Use simultaneous equations to determine the equation of the cubic graph
b
y
y
(−4, 0)
x
(2, 24)
(5, 0)
x
0 (0, 0)
c
PA
G
E
(−1, 0)
0
PR
O
(0, 16)
(−8, 0)
FS
a
O
Consolidate
containing the points (0, 3), (1, 4), (−1, 8), (−2, 7).
3 WE16 Solve the inequations.
a (x − 2) 2 (6 − x) > 0
b 5 x : 4x ≤ x3 6
c 2(x + 4) 3 − 16 < 0
4 Calculate 5 x : 3x3 + 7 > 7x2 + 3x 6 .
5 WE17 Sketch the graphs of y = (x + 2)(x − 1) 2 and y = −3x and calculate
the coordinates of the points of intersection. Hence state the values of x
for which −3x < (x + 2)(x − 1) 2.
6 Calculate the coordinates of the points of intersection of y = 4 − x2 and
y = 4x − x3 and then sketch the graphs on the same axes.
7 Determine the equation for each of the following graphs of cubic polynomials.
d
y
(1, 0)
0
(5, 0)
x
x
(0, −20)
EC
(0, 0)
0
TE
D
(1, –3) stationary
point of inflection
y
8 a
Give the equation of the graph which has the same shape as y = −2x3 and a
point of inflection at (−6, −7).
R
b Calculate the y-intercept of the graph which is created by translating the graph
U
N
C
O
R
of y = x3 two units to the right and four units down.
c A cubic graph has a stationary point of inflection at (−5, 2) and a y-intercept of
(0, −23). Calculate its exact x-intercept.
d A curve has the equation y = ax3 + b and contains the points (1, 3) and
(−2, 39). Calculate the coordinates of its stationary point of inflection.
y
9 The graph of y = P(x) shown is the reflection of
a monic cubic polynomial in the x-axis. The
graph touches the x-axis at x = a, a < 0 and
cuts it at x = b, b > 0.
x
0
a
a Form an expression for the equation of
b
the graph.
b Use the graph to find 5 x : P(x) ≥ 0 6 .
c How far horizontally to the left would the
graph need to be moved so that both of its x-intercepts are negative?
d How far horizontally to the right would the graph need to be moved so that both
of its x-intercepts are positive?
200 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 200
21/07/15 12:02 pm
10 A graph of a cubic polynomial with equation y = x3 + ax2 + bx + 9 has a turning
PA
G
E
PR
O
O
FS
point at (3, 0).
a State the factor of the equation with greatest multiplicity.
b Determine the other x-intercept.
c Calculate the values of a and b.
11 Solve the cubic inequations.
a (x − 2)(x + 1)(x + 9) ≥ 0
b x2 − 5x3 < 0
c 8(x − 2) 3 − 1 > 0
d x3 + x ≤ 2x2
e 5x3 + 6x2 − 20x − 24 < 0
f 2(x + 1) − 8(x + 1) 3 < 0
12 Find the coordinates of the points of intersection of the following.
a y = 2x3 and y = x2
b y = 2x3 and y = x − 1
c Illustrate the answers to parts a and b with a graph.
d Solve the inequation 2x3 − x2 ≤ 0 algebraically and explain how you could
use your graph from part c to solve this inequation.
13 aThe number of solutions to the equation x3 + 2x − 5 = 0 can be found by
determining the number of intersections of the graphs of y = x3 and a
straight line. What is the equation of this line and how many solutions does
x3 + 2x − 5 = 0 have?
b Use a graph of a cubic and a linear polynomial to determine the number of
solutions to the equation x3 + 3x2 − 4x = 0.
c Use a graph of a cubic and a quadratic polynomial to determine the number of
solutions to the equation x3 + 3x2 − 4x = 0.
d Solve the equation x3 + 3x2 − 4x = 0.
TE
D
14 The graph of a polynomial of degree 3 cuts the x-axis at x = 1 and at x = 2. It
U
N
C
O
R
R
EC
cuts the y-axis at y = 12.
a Explain why this is insufficient information to completely determine the
equation of the polynomial.
b Show that this information identifies a family of cubic polynomials with
equation y = ax3 + (6 − 3a)x2 + (2a − 18)x + 12.
c On the same graph, sketch the two curves in the family for which a = 1
and a = −1.
d Determine the equation of the curve for which the coefficient of x2 is 15.
Specify the x-intercepts and sketch this curve.
15 The graph with equation y = (x + a) 3 + b passes through the three points
(0, 0), (1, 7), (2, 26).
a Use this information to determine the values of a and b.
b Find the points of intersection of the graph with the line y = x.
c Sketch both graphs in part b on the same axes.
d Hence, with the aid of the graphs, find 5 x : x3 + 3x2 + 2x > 0 6 .
16 aShow the line y = 3x + 2 is a tangent to the curve y = x3 at the point
(−1, −1).
b What are the coordinates of the point where the line cuts the curve?
c Sketch the curve and its tangent on the same axes.
d Investigate for what values of m will the line y = mx + 2 have one, two or three
intersections with the curve y = x3.
Topic 4 Cubic polynomials c04CubicPolynomials.indd 201
201
21/07/15 12:02 pm
MAstEr
17 Give the equation of the cubic graph containing the points (−2, 53), (−1, −6),
(2, 33), (4, −121).
18 a Write down, to 2 decimal places, the coordinates of the points of
Units 1 & 2
Topic 3
Concept 5
O
PR
O
E
A rectangular storage container is
designed to have an open top and a
square base.
U
N
C
18
O
R
R
Cubic models
and applications
Concept summary
WORKeD
eXaMpLe
Practical situations which use cubic polynomials as models are likely to require
a restriction of the possible values the variable may take. This is called a domain
restriction. The domain is the set of possible values of the variable that the polynomial
may take. We shall look more closely at domains in later chapters.
The polynomial model should be expressed in terms of one variable.
Applications of cubic models where a maximum or minimum value of the model is
sought will require identification of turning point coordinates. In a later chapter we
will see how this is done. For now, obtaining turning points may require the use of
graphing or CAS technology.
EC
AOS 1
Cubic models and applications
TE
D
4.6
PA
G
omar Khayyăm (1050–1123) is not only
a brilliant poet, but also an extraordinary
mathematician, and is noted for linking algebra
with geometry by solving cubic equations as the
intersection of two curves.
FS
intersection of y = (x + 1) 3 and y = 4x + 3.
b Form the cubic equation
ax3 + bx2 + cx + d = 0 for which
the x-coordinates of the points of
intersection obtained in part a are
the solution.
c What feature of the graph of
y = ax3 + bx2 + cx + d would the
x-coordinates of these points of
intersection be?
The base has side length x cm and the
height of the container is h cm. The sum
of its dimensions (the sum of the length,
width and height) is 48 cm.
a Express h in terms of x.
h
b Show that the volume V cm3
of the container is given by
V = 48x2 − 2x3.
202
x
x
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 202
21/07/15 12:02 pm
c State any restrictions on the values x can take.
d Sketch the graph of V against x for appropriate values of x, given its
maximum turning point has coordinates (16, 4096).
e Calculate the dimensions of the container with the greatest possible
volume.
THINK
WRITE
a Write the given information
a Sum of dimensions is 48 cm.
b Use the result from part a to b The formula for volume of a cuboid is
V = lwh
∴ V = x2h
Substitute h = 48 − 2x.
PR
O
express the volume in terms
of one variable and prove
the required statement.
FS
x + x + h = 48
h = 48 − 2x
O
as an equation connecting
the two variables.
E
V = x2 (48 − 2x)
∴ V = 48x2 − 2x3, as required
c Length cannot be negative, so x ≥ 0.
d Draw the cubic graph but
48 − 2x ≥ 0
−2x ≥ −48
∴ x ≤ 24
Hence the restriction is 0 ≤ x ≤ 24.
d V = 48x2 − 2x3
U
N
C
O
R
R
EC
only show the section of
the graph for which the
restriction applies. Label the
axes with the appropriate
symbols and label the given
turning point.
Height cannot be negative, so h ≥ 0.
PA
G
Note: It could be argued
that the restriction is
0 < x < 24 because when
x = 0 or x = 48 there is no
storage container, but we
are adopting the closed
convention.
TE
D
c State the restrictions.
= 2x2 (24 − x)
x-intercepts: let V = 0
x2 = 0 or 24 − x = 0
∴ x = 0 (touch), x = 24 (cut)
(0, 0), (24, 0) are the x-intercepts.
This is a negative cubic.
Maximum turning point (16, 4096)
Draw the section for which 0 ≤ x ≤ 24.
V
(16, 4096)
4000
2000
0
2
4
6
8
10
12
14
16
18
20
22
24
x
Topic 4 Cubic polynomials c04CubicPolynomials.indd 203
203
21/07/15 12:02 pm
e 1 Calculate the required
e The maximum turning point is (16, 4096). This means the
greatest volume is 4096 cm3. It occurs when x = 16.
∴ h = 48 − 2(16) ⇒ h = 16
Dimensions: length = 16 cm, width = 16 cm, height = 16 cm
dimensions.
Note: The maximum
turning point (x, V ) gives
the maximum value of V
and the value of x when
this maximum occurs.
The container has the greatest volume when it is a cube of
edge 16 cm.
FS
2 State the answer.
A rectangular storage container
is designed to have an open top and a
square base.
The base has side length x metres and the
height of the container is h metres. The total
length of its 12 edges is 6 metres.
a Express h in terms of x.
b Show that the volume V m3 of the
h
container is given by V = 1.5x2 − 2x3.
x
c State any restrictions on the values x
can take.
d Sketch the graph of V against x for
x
appropriate values of x, given its
maximum turning point has coordinates (0.5, 0.125).
e Calculate the dimensions of the container with the greatest possible volume.
2 A rectangular box with an open top is to be constructed from a rectangular
sheet of cardboard measuring 20 cm by 12 cm by cutting equal squares of
side length x cm out of the four corners and folding the flaps up.
WE18
PR
O
1
R
EC
TE
D
PA
G
E
PrACtisE
O
ExErCisE 4.6 Cubic models and applications
x
R
x
x
U
N
C
O
x
12 cm
x
x
x
x
20 cm
The box has length l cm, width w cm and volume V cm3.
a Express l and w in terms of x and hence express V in terms of x.
b State any restrictions on the values of x.
204
Maths Quest 11 MatheMatICaL MethODs VCe units 1 and 2
c04CubicPolynomials.indd 204
21/07/15 12:02 pm
c Sketch the graph of V against x for appropriate values of x, given the
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
Consolidate
unrestricted graph would have turning points at x = 2.43 and x = 8.24.
d Calculate the length and width of the box with maximum volume and give this
maximum volume to the nearest whole number.
3 The cost C dollars for an artist to
produce x sculptures by contract is
given by C = x3 + 100x + 2000.
Each sculpture is sold for $500
and as the artist only makes the
sculptures by order, every sculpture
produced will be paid for. However,
too few sales will result in a loss to
the artist.
a Show the artist makes a loss if
only 5 sculptures are produced
and a profit if 6 sculptures are produced.
b Show that the profit, P dollars, from the sale of x sculptures is
given by P = −x3 + 400x − 2000.
c What will happen to the profit if a large number of sculptures are produced?
Why does this effect occur?
d Calculate the profit (or loss) from the sale of:
i 16 sculptures
ii 17 sculptures.
e Use the above information to sketch the graph of the profit P for 0 ≤ x ≤ 20.
Place its intersection with the x-axis between two consecutive integers but don’t
attempt to obtain its actual x-intercepts.
f In order to guarantee a profit is made, how many sculptures should the
artist produce?
4 The number of bacteria in a slow-growing culture at time t hours after 9 am is
given by N = 54 + 23t + t3
a What is the initial number of bacteria at 9 am?
b How long does it take for the initial number of bacteria to double?
c How many bacteria are there by 1 pm?
d Once the number of bacteria reaches 750, the experiment is stopped. At what
time of the day does this happen?
5 Engineers are planning to build an underground tunnel through a city to ease
traffic congestion.
The cross-section of their plan is bounded by the curve shown.
y(km)
O
x(km)
Topic 4 Cubic polynomials c04CubicPolynomials.indd 205
205
21/07/15 12:02 pm
The equation of the bounding curve is y = ax2 (x − b) and all measurements are in
kilometres.
It is planned that the greatest breadth of the bounding curve will be 6 km and the
greatest height will be 1 km above this level at a point 4 km from the origin.
a Determine the equation of the bounding curve.
b If the greatest breadth of the curve was extended to 7 km, what would be the
greatest height of the curve above this new lowest level?
O
7 km
PR
O
O
FS
y
x
E
6 Find the smallest positive integer and the largest negative integer for which the
U
N
C
O
R
R
EC
TE
D
PA
G
difference between the square of 5 more than this number and the cube of 1 more
than the number exceeds 22.
V
7 A tent used by a group of bushwalkers is in the shape
of a square-based right pyramid with a slant height
8m
of 8 metres.
P
For the figure shown, let OV, the height of the tent, be
h metres and the edge of the square base be 2x metres.
2x m
O
a Use Pythagoras’ theorem to express the length
M
of the diagonal of the square base of the tent in
N
2x m
terms of x.
b Use Pythagoras’ theorem to show 2x2 = 64 − h2.
1
c The volume V of a pyramid is found using the formula V = 3Ah where A is the
area of the base of the pyramid. Use this formula to show that the volume of
space contained within the bushwalkers’ tent is given by V = 13 (128h − 2h3).
diIf the height of the tent is 3 metres, what is the volume?
ii
What values for the height does this mathematical model allow?
1
e Sketch the graph of V = (128h − 2h3) for appropriate values of h and estimate
3
the height for which the volume is greatest.
f The greatest volume is found to occur when the height is half the length of
the base. Use this information to calculate the height which gives the greatest
volume and compare this value with your estimate from your graph in part e.
8 A cylindrical storage container is designed so that it is open at the top and has a
surface area of 400π cm2. Its height is h cm and its radius is r cm.
400 − r2
a Show that h =
.
2r
1
b Show that the volume V cm3 the container can hold is given by V = 200πr − 2πr3.
c State any restrictions on the values r can take.
206 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 206
21/07/15 12:02 pm
d Sketch the graph of V against r for
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
appropriate values of r.
e Find the radius and height of the
container if the volume is 396π cm3.
f State the maximum possible
volume to the nearest cm2 if the
maximum turning point on the
1
graph of y = 200πx − πx3 has an
2
20
x-coordinate of
.
!3
y
9 A new playground slide for children is to be
constructed at a local park. At the foot of the
slide the children climb a vertical ladder to
reach the start of the slide. The slide must
(2.1)
start at a height of 2.1 metres above the
ground and end at a point 0.1 metres above
the ground and 4 metres horizontally
from its foot. A model for the slide is
(4, 0.1)
h = ax3 + bx2 + cx + d where h metres is
x
0
the height of the slide above ground level at a
1
2
3
4
5
horizontal distance of x metres from its foot.
The foot is at the origin.
The ladder supports the slide at one end and the slide also requires two vertical
struts as support. One strut of length 1 metre is placed at a point 1.25 metres
horizontally from the foot of the slide and the other
is placed at a point 1.5 metres horizontally from
the end of the slide and is of length 1.1 metres.
a Give the coordinates of 4 points which lie on
the cubic graph of the slide.
b State the value of d in the equation of the slide.
c Form a system of 3 simultaneous equations, the
solutions to which give the coefficients a, b, c in
the equation of the slide.
d The equation of the slide can be
shown to be y = −0.164x3 + x2 − 1.872x + 2.1.
Use this equation to calculate the length of a third strut thought necessary at
x = 3.5. Give your answer to 2 decimal places.
10 Since 1988, the world record times for the
men’s 100-m sprint can be roughly
approximated by the cubic model
T(t) = −0.00005(t − 6) 3 + 9.85 where T
is the time in seconds and t is the number
of years since 1988.
a In 1991 the world record was 9.86
seconds and in 2008 the record was
9.72 seconds. Compare these times with
those predicted by the cubic model.
topic 4 CubIC pOLynOMIaLs
c04CubicPolynomials.indd 207
207
21/07/15 12:02 pm
b Sketch the graph of T versus t from 1988 to 2008.
c What does the model predict for 2016? Is the model likely to be a good
U
N
O
C
Master
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
predictor beyond 2016?
y
11 A rectangle is inscribed under the parabola
2
y = 9 − (x − 3) so that two of its corners
lie on the parabola and the other two lie
on the x-axis at equal distances from the
(x, y)
intercepts the parabola makes with the x-axis.
a Calculate the x-intercepts of the parabola.
x
0
b Express the length and width of the
rectangle in terms of x.
c Hence show that the area of the rectangle
y = 9 – (x – 3)2
3
2
is given by A = −2x + 18x − 36x.
d For what values of x is this a valid model of the area?
e Calculate the value(s) of x for which A = 16.
12 A pathway through the countryside
passes through 5 scenic points. Relative
to a fixed origin, these points have
coordinates A(−3, 0), B(−!3, −12 !3),
C(!3, 12 !3), D(3, 0); the fifth
scenic point is the origin, O(0, 0).
The two-dimensional shape of the path
is a cubic polynomial.
a State the maximum number of turning
points and x-intercepts that a cubic graph can have.
b Determine the equation of the pathway through the 5 scenic points.
c Sketch the path, given that points B and C are turning points of the cubic
polynomial graph.
d It is proposed that another pathway be created to link B and C by a direct route.
Show that if a straight-line path connecting B and C is created, it will pass
through O and give the equation of this line.
e An alternative plan is to link B and C by a cubic path which has a stationary
point of inflection at O. Determine the equation of this path.
Use CAS technology to answer the following questions.
13 a Consider question 7e.
1
Use the calculator to sketch the graph of V = (128h − 2h3) and state the
3
height for which the volume is greatest, correct to 2 decimal places.
b Consider question 7f.
Show that the greatest volume occurs when the height is half the length
of the base.
c Consider question 8f.
Use technology to obtain the maximum volume and calculate the corresponding
dimensions of the cylindrical storage container, expressed to 1 decimal place.
14 a Consider question 9c.
Use technology to obtain the equation of the slide.
b Consider question 11.
Calculate, to 3 decimal places, the length and width of the rectangle which has
the greatest area.
208 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 208
21/07/15 12:02 pm
4.7 Review
A summary of the key points covered in this topic is
also available as a digital document.
REVIEW QUESTIONS
Download the Review questions document from
the links found in the Resources section of your
eBookPLUS.
Activities
to access eBookPlUs activities, log on to
TE
D
www.jacplus.com.au
PA
G
ONLINE ONLY
E
PR
O
the review contains:
• short-answer questions — providing you with the
opportunity to demonstrate the skills you have
developed to efficiently answer questions without the
use of CAS technology
• Multiple-choice questions — providing you with the
opportunity to practise answering questions using
CAS technology
• Extended-response questions — providing you with
the opportunity to practise exam-style questions.
FS
the Maths Quest review is available in a customisable
format for you to demonstrate your knowledge of this
topic.
www.jacplus.com.au
O
ONLINE ONLY
Interactivities
Units 1 & 2
Cubic polynomials
Sit topic test
U
N
C
O
R
R
EC
A comprehensive set of relevant interactivities
to bring difficult mathematical concepts to life
can be found in the Resources section of your
eBookPLUS.
studyON is an interactive and highly visual online
tool that helps you to clearly identify strengths
and weaknesses prior to your exams. You can
then confidently target areas of greatest need,
enabling you to achieve your best results.
topic 4 CubIC pOLynOMIaLs
c04CubicPolynomials.indd 209
209
21/07/15 12:02 pm
4 Answers
16a k = 25
1 A: Degree 5; leading coefficient 4; constant term 12;
17a a = 3; b = −2; c = −8
b m = −3; n = −10; p = 24
c a = 1; b = 7; c = −41; x2 − 14x + 8 = (x − 7) 2 − 41
d 4x3 + 2x2 − 7x + 1
O
iii m + n
19a a = −3; b = −1; p = 4; q = −2
b i 4x4 + 12x3 + 13x2 + 6x + 1 = (2x2 + 3x + 1) 2
ii 2x2 + 3x + 1
20a k = −13; m = −5; n = 6
b (x − 3) 2 (x − 2) 2 (x + 3)(x + 2)
3x2 − 6x + 5
29
= 3x − 12 +
;
x+2
x+2
P(x) = 3x − 12, d = 29
21
5
Q
D
4
F
6
3x5
12
R
−5x4
 9
Z
−18x4
 0
N
49x6
O
4
U
N
B
Constant
term
C
A
Leading
term
R
Type of
coefficient
R
EC
TE
D
2x3 − 5x2 + 8x + 6
18
11a
= 2x2 − x + 6 +
; quotient
x−2
x−2
is 2x2 − x + 6; remainder is 18.
x3 + 10
81
1
1
1
b
= − x2 − x − +
;
1 − 2x
2
4
8 8(1 − 2x)
81
remainder is .
8
12Quotient is x2 − x + 3; remainder is 0.
13a A, B, D, F are polynomials.
Degree
 9
1
2
b C is not a polynomial due to "4x5 = 2x term.
E is not a polynomial due to
14a 78
5
5
= x−2 term.
2
3
3x
b −12
c 0
d −6
e −6
f −5.868
15a −14a
b h2 − 5h − 4
c 2xh + h2 − 7h
210 FS
= 4x2 (x + 1) − 2x(x + 1) − 5(x + 1) + 6
18a i 3x4 + 7x + 12
ii 6x4 + 10x2 − 21x − 31
iii 6x6 − 21x5 − 29x4 − 14x3 − 20x2 − 7x − 11
b i m
ii m
E
10a = 3, b = −12, c = 29;
d b = 4; c = 5
PA
G
coefficients ∈ Z
C: Degree 2; leading coefficient −0.2; constant term 5.6;
coefficients ∈ Q
2 y7 + 2y5 − !2y2 + 4 (answers may vary)
3a 15
b 10
4 21
5 a = 2; b = −13; c = 6
6 (x + 2) 3 = x2 (x + 1) + 5x(x + 2) + 2(x + 3) + 2
7 p = 5; q = −4
8 x7 + 4x5 − 6x4 + 5x3 − 4x2 − 5x + 2; degree 7
x − 12
15
9a
=1−
; quotient is 1; remainder is −15.
x+3
x+3
4x + 7
5
b
=2+
2x + 1
2x + 1
b a = −9
1
c n = −
5
PR
O
Exercise 4.2
Quotient
Remainder
a
1
   9
b
4
    1
c
x+7
−10
d
2x − 12
  27
e
x2
f
x2 − 7x + 2
+ 5x + 12
  41
   0
18
1 + 2x
20
2
b 2x + 3x + 5 +
2x − 3
48
c x2 + 6 −
x+6
1
2
d x − x + 1 +
x+1
22a 4x2 + x − 3 +
e x2 + x +
3x + 5
x2 − 1
7x + 12
f −2x − 2 −
(x + 2)(x − 3)
23a 2x2 −
139
49
13
x−
+
2
4(2x + 3)
4
139
49
13
; quotient is 2x2 − x + .
4
2
4
139
c Dividend equals −
.
4
d Divisor equals 0.
b Remainder is −
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 210
21/07/15 12:03 pm
24a Define using CAS technology.
c i P(x) = (x − 4) 3; Q(x) = (x − 4)(x2 + 4x + 16)
349
9
c 24n2 + 24an − 16n − 154
d 9
ii Proof required
b
d Third factor is (x − 3); b = c = −3; d = 9
Exercise 4.3
1a The remainder is 19.
b a = 1; a = 2
e 0
f 26
10a–10
b 8
2
c
d
e
14a (x − 1)(x + 2)(x + 4)
U
N
b (x + 2)(x + 3)(x + 5)
c (x − 2)(2x + 1)(x − 5)
d (x + 1)(3x − 4)(1 − 6x)
e (x − 1)(x + 3)(x − 2)
f (x + 1)(x − 7)(x + 7)
15a (2x + 1)(3x − 1)(4x + 5)
y-intercept
x-intercept
(1, −8)
(0, −9)
(3, 0)
b
(−6, 1)
(0, −5)
(−2.7, 0) approx.
a
y = (x – 1)3 – 8
y
EC
C
f
Inflection point
a
(3, 0)
R
b
PR
O
TE
D
b m = −3; n = −3
x
0
(1, –8)
R
13a
1 (0, –9)
O
b
Exercise 4.4
d –19
−
+ 17x + 90
i (x + 4)(x + 1)(1 − 2x)
ii −2x3 − 9x2 − 3x + 4
(x − 4)(x + 1)(x + 2)
(x + 12)(3x + 1)(x + 1)
(5x + 1)(2x + 3)(2x + 1)
(4x − 3)(5 − 2x)(5 + 2x)
−(x − 3) 2 (8x − 11)
(3x − 5) 2 (x − 5)
Q(x) = (x − 3)(x + 3)(x + 1)
d x3 + 7x2 + 15x + 9 : x = −3, x = −1;
x3 − 9x2 + 15x + 25 : x = 5, x = −1
19−12!2 + 33
20−0.4696
E
d −10 8
c a = −3; b = 1; P(x) = (x − 3)(x2 + 1),
PA
G
3
c –101
12x2
1
2
18a (x − 2)(x + 2)(5x + 9); k = 9
P(x) = (x + 1)(x + 5)(x − 3)
b P(x) = (x + 1)(3x + 2)(4x + 7)
6 Linear factors are (2x − 1), (2x + 1) and (3x + 2).
1 1
7 x = −2, , −
3 2
3
8 x = 0, x = − , x = 2, x = −2
2
9a–5
b 2
12a i (x − 5)(x − 9)(x + 2)
f 0, 1, − , –20
c (2x − 1)(x − 5) 2; equation has roots x = , x = 5
5a
P (−1) = 0 ⇒ (x + 1) is a factor;
11a a = 2; b = 3
3
4
2
3
b Proof required
4 a = 0; a = −1
c –19
3
2
17a Proof required; (x − 2) 1 x + 4 − !7 2 1 x + 4 + !7 2
b a = −9; b = 8; P(x) = 3x3 − 9x2 + 8x − 2
ii
d 1, − , –3
e 1, −
3a Proof required to show Q(2) = 0
x3
c –1, 6, 8
FS
2 k = 2
37
.
8
b 7, − , 9
O
b The remainder is
5
3
16a −4, 3, −5
b
y
1
y = 1 – – (x + 6)3
36
(–6, 1) (–2.7, 0)
x
0
(0, –5)
b i (2x − 5)
ii (2x − 5)(4x + 1)(x − 1)
iii m = −26
Topic 4 Cubic polynomials c04CubicPolynomials.indd 211
211
21/07/15 12:03 pm
2 Inflection point
y-intercept
x-intercept
a
(6, 0)
(0, −27)
(6, 0)
b
(0, −2)
(0, −2)
(1, 0)
a
5 y-intercept
x-intercepts
a
(0, 6)
(−6, 0) and (3, 0) which is a
turning point
b
(0, 8)
(−2, 0) is a turning point and (1, 0)
y
(
x –3
y = ––
2
)
a
3
y
1 (x – 3)2(x + 6)
y=–
9
(6, 0)
(0, 6)
(–6, 0)
x
0
(3, 0)
x
FS
0
(0, –27)
y = 2x3 – 2
(0, 6)
(–2, 0)
x
0
PR
O
(1, 0)
(1, 0)
0
(0, −24)
(−6, 0), (−1, 0), (4, 0)
1
Q− , 0R,
2
(0, −24)
b
a
(2, 0), (4, 0)
y
y = (x + 1)(x + 6)(x – 4)
(–6, 0)
(–1, 0)
0
(4, 0)
EC
x
R
y
b
(4, 0)
O
– 1– , 0
2
R
y = (x – 4)(2x + 1)(6 – x)
( )
(0, 0)
U
N
4 x-intercept
(0, 0) and (10, 0) is a turning point
(10, 0)
(0, 0)
0
x
y ≤ 0.1x (10 – x)2
7 x3 − 3x2 − 10x + 24 = (x − 2)(x − 4)(x + 3)
y-intercept
x-intercepts
(0, 24)
(−3, 0), (2, 0), (4, 0)
(6, 0)
x
y = x3 – 3x2 – 10x + 24
(0, 24)
y-intercept
x-intercepts
(0, 0)
x
y
C
0
y-intercept
TE
D
a
6
x-intercepts
PA
G
y-intercept
E
(0, –2)
3 O
y y = –2(x – 1)(x + 2)2
b
y
b
(–3, 0)
(2, 0)
(4, 0)
0
(−2, 0), (0, 0), (2, 0)
y
y = 3x(x2 – 4)
(–2, 0)
(0, 0) (2, 0)
0
x
8a −x3 + 3x2 + 10x − 30 = −(x − 3) Q x − !10 R
1 x + !10 2
y-intercept
(0, −30)
212 x-intercepts
1 −!10, 0 2 , (3, 0), 1 !10, 0 2
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 212
21/07/15 1:38 pm
y
– 10
a
10
3
y
x
0
(0, 37)
y = –x3 + 3x2 + 10x – 30
(0, –30)
(–1, 0)
x
0
(–4, –27)
b Stationary point of inflection (−1, 1); y-intercept (0, 2);
y
y = x3 + 3x2 + 3x + 2
(1 – 3 5, 0)
(−2, 0)
(1, 10)
PR
O
x
0
x
y
EC
y = –x3
3
(
3,0
3––
3
2
x
0
(0, –27)
y
d
x
R
0
(–2, 16)
(0, 0)
x
0
y=
–x3 +
3
U
N
C
O
(
y = –3x3
R
(−3, 0)
(3, 27)
TE
D
y = 3x3
b y = –(x + 3)3
y
PA
G
y = x3
0
E
c
3
y = x3 + 3
x
0
y
y = (x + 3)3
O
(0, 8)
(0, 2)
(−1, 1)
9a
y
b
FS
x-intercept (−2, 0)
10
Inflection point
y-intercept
x-intercept
a
(−4, −27)
(0, 37)
(−1, 0)
b
(1, 10)
(0, 8)
(−0.7, 0) approx.
c
(3, 27)
(0, −27)
(0.6, 0) approx.
d
(−2, 16)
(0, 0)
(0, 0)
e
Q−
f
4
, 0R
3
(0, 9)
(0, −48)
(0, 9)
4
, 0R
3
Q−
(−3, 0)
e
y
( (
4, 0
––
3
x
0
(0, –48)
Topic 4 Cubic polynomials c04CubicPolynomials.indd 213
213
21/07/15 12:03 pm
d
y
(0, 9)
x
y-intercept
x-intercepts
a
(0, −8)
b
(0, 0)
(−8, 0), (0, 0), (5, 0)
c
(0, −12)
(−3, 0), (1, 0), (4, 0)
d
(0, 12)
(−4, 0), (2, 0), (6, 0)
e
(0, 7)
f
Q 0, R
Q−
e
y
y = 0.1(2x – 7)(x –10)(4x + 1)
(−4, 0), (−1, 0), (2, 0)
1
, 0 R , Q 72, 0 R ,
4
5
2
8
, 0 R , Q 58, 0 R ,
3
( –27 , 0
(0, 7)
(– –14 , 0
(2, 0)
y
f
PA
G
(2, 0)
(–1, 0)
x
0
TE
D
(0, –8)
12
(0, 0)
0
(5, 0)
x
U
N
c
C
O
R
R
(–8, 0)
y = –0.5x(x + 8)(x – 5)
EC
y
b
(
2.5
0
(–85 , 0
y-intercept
(2, 0)
x
x-intercepts
(0, 32)
(−4, 0) is a turning point;
(2, 0) is a cut
b
(0, 54)
(3, 0) is a turning point;
(−3, 0) is a cut
c
(0, 36)
(−3, 0) is a turning point;
(4, 0) is a cut
d
(0, −12)
(2, 0) is a turning point;
(12, 0) is a cut
e
(0, 0)
Q−
y = (x + 3)(x – 1)(4 – x)
(1, 0) (4, 0)
(
a
3
, 0R
2
is a turning point;
(0, 0) is a cut
f
(–3, 0)
(
5
x – 1 3x
–+2 x– –
y=2 –
4
8
2
E
y = (x – 2)(x + 1)(x + 4)
(– –38 , 0
(–4, 0)
(10, 0)
x
0
(10, 0)
Q−
y
(6, 0)
0
FS
0
a
(0, 12)
(2, 0)
(–4, 0)
PR
O
(–3, 0)
11
1 (2 – x)(6 – x)(4 + x)
y=–
4
O
f
(0, 0)
(0, 0) is a turning point;
(0.4, 0) is a cut
0
(0, –12)
214 Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 214
21/07/15 1:39 pm
13
(0, 32)
y = – (x + 4)2(x – 2)
(2, 0)
0
(−4, 0)
b
y = 2(x + 3)(x – 3)2
Stationary
point of
inflection
y-intercept
x-intercepts
a
(−3, 0)
(0, 27)
(−3, 0)
b
none
(0, −9)
c
none
(0, −15)
(−3, 0) is a turning
point; Q 12, 0 R is a cut
d
(0, 54)
e
Q
1
, 1R
2
Q 0,
none
(−3, 0), Q 12, 0 R , (5, 0)
3
B
2
(1.1, 0) approx.
(0, 0)
Q
1
, 0R
4
is a turning
Q
2
, 0R
3
is a turning
FS
a
point; (0, 0) is a cut
f
(3, 0)
(–3, 0)
none
(0, 4)
O
0
PR
O
a
y
c
y
(0, 27)
y = (x + 3)3
y = (x + 3)2(4 – x)
(0, 36)
0
y
(12, 0)
x
EC
(0, –12)
y
– 3– , 0
2
y
y = (x + 3)2(2x – 1)
1– , 0
2
( )
(–3, 0)
0
c
y
y = 3x (2x + 3)2
)
R
(
TE
D
1
y = – (2x – x)2(x – 12)
4
e
x
0
b
(2, 0)
0
PA
G
(4, 0)
x
E
(–3, 0)
(–3, 0)
d
(0, 0)
(–3, 0)
x
0
x
(0, –9)
y = (x + 3)(2x – 1)(5 – x)
( 1–2 , 0)
(5, 0)
(0, –15)
x
U
N
C
O
R
0
point; Q −23, 0 R is a cut
f
d
y
2(y – 1) = (1 – 2x)3
y
(0, 0)
0
y = –0.25x2(2 – 5x)
(0.4, 0)
x
( ,1)
(0, )
1
–
2
3
–
2
(1 + 3 2, 0)
0
x
Topic 4 Cubic polynomials c04CubicPolynomials.indd 215
215
21/07/15 12:03 pm
e
y
c
4y = x(4x – 1)2
y = 9x2 – 3x3 + x – 3
( )
1– , 0
4
(0, 0)
0
( 33, 0)
x
(–
y
f
y=–
)
3, 0
3
0
y
d
1
(2 – 3x)(3x + 2)(3x – 2)
2
x
(3, 0)
(0, –3)
(0, 4)
0
( 2–3 , 0)
(–3, 0)
0
(–1, 0)
(0, 0)
x
e
14See table at foot of page.*
a
y
PR
O
O
(– 2–3 , 0)
FS
y = 9x(x2 + 4x + 3)
x
y
y = 9x2 – 2x3
(0, 9)
y = 9x3 + 27x2 + 27x + 9
E
(–1, 0)
x
PA
G
( 9–2 , 0)
0 (0, 0)
y
f
(0, 9)
y = 9x3 – 4x
(0, 0)
(2–3 , 0)
0
(–1, 0)
0
(1, 0)
x
x
216 Factorised form
a
y=
b
y = x(3x − 2)(3x + 2)
c
d
y = −3(x − 3) ax −
x2 (9
− 2x)
R
*14
R
EC
(– 2–3 , 0)
y = –9x3 – 9x2 + 9x + 9
TE
D
y
b
x
0
Stationary point of inflection
y-intercept
x-intercepts
(0, 0)
none
(0, 0)
none
(0, −3)
y = 9x(x + 1)(x + 3)
none
(0, 0)
(−3, 0), (−1, 0), (0, 0)
e
y = 9(x + 1) 3
(−1, 0)
(0, 9)
(−1, 0)
f
y = −9(x +
none
(0, 9)
(−1, 0) is a turning point;
1 1, 0 2 is a cut
U
N
C
O
none
1) 2 (x
!3
!3
b ax +
b
3
3
− 1)
(0, 0) is a turning point;
9
a , 0b is a cut
2
2
2
a− , 0b, (0, 0), a , 0b
3
3
a−
!3
!3
, 0b, a
, 0b, (3, 0)
3
3
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 216
21/07/15 1:39 pm
15See table at foot of page.*
y=
y
(– 3–2 , 0) (–1, 0)
3x2 –
y
17x – 12
(– –13 , 0)
x
0 (0, –18)
x
(0, –12)
y
(–92 , 0)
(–2, 0)
(4, 0)
0
b
y = 6x3 – 13x2 – 59x – 18
e
y = 6 – 55x + 57x2 – 8x3
y
FS
a
d
2x3 –
y = –5x3 – 7x2 + 10x + 14
(– –72 , 0)
PR
O
(– 2, 0)
O
(0, 14)
( 2, 0)
x
0
( 1–8 , 0)
(6, 0)
(1, 0)
x
0
c
PA
G
f
y
(–6, 0)
0
1 x2 + 14x – 24
y = ––
2
(4, 0)
(2, 0)
x
(0, –24)
(4, 0)
x
Factorised form
R
*15
R
EC
(0, 4) (–2 + 15, 0)
0
y
TE
D
y = x3 – 17x + 4
(–2 – 15, 0)
E
(0, 6)
y = (x + 1)(2x + 3)(x − 4)
b
y = −(x − 1)(8x − 1)(x − 6)
U
N
C
O
a
c
d
e
f
y = (x − 4) 1 x + 2 − !5 2 1 x + 2 + !5 2
y = (x + 2)(3x + 1)(2x − 9)
y = (5x + 7) 1 !2 − x 2 1 !2 + x 2
1
y = − (x − 2)(x + 6)(x − 4)
2
y-intercept
(0, −12)
(0, 6)
(0, 4)
(0, −18)
(0, 14)
(0, −24)
x-intercepts
3
a− , 0b, (−1, 0), (4, 0)
2
1
a , 0b, (1, 0), (6, 0)
8
1 −2 − !5, 0 2 , 1 −2 + !5, 0 2 , (4, 0)
9
1
(−2, 0), a− , 0b, a , 0b
3
2
7
(−!2, 0), a− , 0b, (!2, 0)
2
(−6, 0), (2, 0), (4, 0)
Topic 4 Cubic polynomials c04CubicPolynomials.indd 217
217
21/07/15 1:39 pm
16a k = −19
y = x3 + 4x2 – 44x – 96
y
b P(x) = (3x − 1)(5x + 1)(2x − 1)
(–5.4, 100.8)
1 1 1
5 3 2
1
1
1
d y-intercept (0, 1); x-intercepts a− , 0b, a , 0b, a , 0b
5
3
2
c x = − , ,
0
x
(0, –96)
y
(2.7, –166.0)
b Proof required
1
–
3
0
20See table at foot of page.*
x
1
–
2
a
y = 10x3 – 20x3 – 10x – 19
y
y ≥ 30x3 – 19x2 + 1
(–0.2, –17.9)
f Shade the closed region above the graph of y = P(x).
−
4) 3
b
E
y = x3 – 5x2 + 11x – 7
2
(
x
12, 0
TE
D
(4 +
0
Stationary point of inflection (4, 6); y-intercept (0, 38);
x-intercept approximately (6.3, 0)
18aProof required — check with your
c
EC
teacher. y = (x − 1)(x2 − 4x + 7)
b Proof required — check with your teacher.
c y → ∞
d
R
O
y (0.2, 502.3)
y = 9x3 – 70x2 + 25x + 500
(5, 0)
x
1
(x − 3) 3 − 7
49
b y = 0.25x(x + 5)(x − 4)
c y = −(x + 2) 2 (x − 3)
2 y = 2x3 + 3x2 − 4x + 3
3a x < 6, x ≠ 2
b 5 x : −2 ≤ x ≤ 0 6 ∪ 5 x : x ≥ 2 6
1a y =
x
U
N
19aMaximum turning point (−5.4, 100.8); minimum
turning point (2.7, −166.0); y-intercept (0, −96);
x-intercepts (−8, 0), (−2, 0), (6, 0)
Maximum turning point
(0, 500)
Exercise 4.5
C
0
x
0
R
y
(1, 0)
0
(–2.2, 0)
y = x3 – 5x2 + 11x – 7
218 y = –x3 + 5x2 – 11x + 7
(0, 7)
(4, 6)
*20
y
y
(0, 38)
x
(1.5, – 45.3)
+6
b
(0, –19)
PA
G
17a
−12 (x
0
PR
O
e No
O
(2.6, 0)
FS
1
––
5
c x < −2
Minimum turning point
y-intercept
x-intercept
(1.5, −45.3)
(0, −19)
(2.6, 0)
1 1, 0 2
a
(−0.2, −17.9)
b
none
none
(0, 7)
c
(0.2, 502.3)
(5, 0)
(0, 500)
(−2.2, 0), (5, 0)
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 218
21/07/15 1:40 pm
7
4 5 x : −1 < x < 1 6 ∪ U x : x > 3 V
1 1
B
2 4
12a (0, 0), A ,
3
3
5 Point of intersection is 1 −"2, 3"2 2 .
3
When x > −"2, −3x < (x + 2)(x −
b (−1, −2)
1) 2.
c
y = 2x3
( (
1
1, –
–
(0, 0) 2 4
y = –3x
y = x2
0
2, 3 2)
3
(1, 0)
(0, 0)
(–1, –2)
y = (x + 2)(x – 1)2
x
0
FS
(–
3
y=x–1
x
O
d x ≤ 0.5
13a y = −2x + 5; 1 solution
PR
O
b y = x3 + 3x2, y = 4x; 3 solutions
6 Points of intersection are (1, 3), (2, 0), (−2, 0).
c There are 3 solutions. One method is to use
(0, 4)
(–2, 0)
(1, 3)
x
0
x2
1
7a y = 2 (x + 8)(x + 4)(x + 1)
b y = −2x2 (x − 5)
−3
4
d y = 5 (x − 1)(x − 5) 2
8a y = −2(x + 6) 3 − 7
b (0, −12)
d (0, 7)
R
c Q "10 − 5, 0 R
R
3
EC
c y = −3(x −
1) 3
TE
D
y=4–
O
9a y = −(x − a) 2 (x − b)
b 5 x : x ≤ b 6
c
(2, 0)
y = –3x
a=1
(2, 0)
(–6, 0)
(1, 0)
(0, 12)
(6, 0)
x
0
a = –1
If a = 1, graph contains the points (1, 0), (2, 0),
(0, 12), (−6, 0); if a = −1, the points are
(1, 0), (2, 0), (0, 12), (6, 0).
d Equation is y = −3x3 + 15x2 − 24x + 12 or
y = −3(x − 1)(x − 2) 2; x-intercepts are (1, 0), (2, 0);
(2, 0) is a maximum turning point.
C
c More than b units to the left
y
PA
G
y = 4x – x3
E
y = x3, y = −3x2 + 4x.
d x = −4, 0, 1
14a Fewer than 4 pieces of information are given.
b Proof required — check with your teacher.
y
U
N
d More than −a units to the right
10a (x − 3) 2
(0, 12)
b (−1, 0)
y = –3(x – 1)(x – 2)2
c a = −5, b = 3
(1, 0)
11a −9 ≤ x ≤ −1 or x ≥ 2
b x >
0
1
5
(2, 0)
x
c x > 2.5
d x ≤ 0 or x = 1
6
e x < −2 or −5 < x < 2
3
1
f −2 < x < −1 or x > −2
Topic 4 Cubic polynomials c04CubicPolynomials.indd 219
219
21/07/15 12:18 pm
15a a = 1; b = −1
c
b (−2, −2), (−1, −1), (0, 0)
c
y = (x + 1)3 – 1
V = x (20 – 2x) (12 – 2x)
200
y=x
(0, 0)
0
V
250
150
x
100
(–1, –1)
50
d 5 x : −2 < x < −1 6 ∪ 5 x : x > 0 6
0
16a Proof required — check with your teacher.
c
3
4
5
x
6
x-intercepts at x = 10, x = 6, x = 0 but since 0 ≤ x ≤ 6,
the graph won’t reach x = 10; shape is of a positive
cubic.
d Length 15.14 cm; width 7.14 cm; height 2.43 cm;
greatest volume 263 cm3
3a Loss of $125; profit of $184
b Proof required — check with your teacher.
c Too many and the costs outweigh the revenue from
the sales. A negative cubic tends to −∞ as x becomes
very large.
d i Profit $304
ii Loss $113
PR
O
y = x3
y = 3x + 2
(2, 8)
(0, 0)
x
0
d One intersection if m < 3; two intersections if m = 3;
three intersections if m > 3
17 y = −6x3 + 12x2 + 19x − 5
18a (−3.11, −9.46), (−0.75, 0.02), (0.86, 6.44)
b x3 + 3x2 − x − 2 = 0
c x-intercepts
e
TE
D
P
2000
3 − 4x
2
b Proof required — check with your teacher.
3
c 0 ≤ x ≤ 4
R
O
V
C
0.2
U
N
(0.5, 0.125)
0.1
0.05
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.75
x
x-intercepts at x = 0 (touch), x = 0.75 (cut); shape of a
negative cubic
e Cube of edge 0.5 m
2a l = 20 − 2x; w = 12 − 2x; V = (20 − 2x)(12 − 2x)x
b 0 ≤ x ≤ 6
220 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
–2000
R
1a h =
0.15
0
EC
Exercise 4.6
d
PA
G
E
(–1, –1)
2
O
b (2, 8)
1
FS
(–2, –2)
x
(20, –2000)
x-intercepts lie between 5 and 6 and between
16 and 17.
f Between 6 and 16
4a 54
b 2 hours
c 210
d 5 pm
1
5a y = − x2 (x − 6)
32
81
b
km
32
6 −4, 1
7a 2!2x
b Proof required — check with your teacher.
c Proof required — check with your teacher.
d i 110 m3
ii Mathematically 0 ≤ h ≤ 8
Maths Quest 11 MATHEMATICAL METHODS VCE Units 1 and 2
c04CubicPolynomials.indd 220
21/07/15 12:03 pm
e
11a x = 0, x = 6
1 (128h – 2h3)
V=–
3
V
110
b Length 2x − 6; width 6x − x2
(3, 110)
c Proof required
d 3 ≤ x ≤ 6
8a
b
c
d V
b y = −2x(x2 − 9)
y
c
C( 3, 12 3)
A(–3, 0)
B(– 3, 12 3)
d y = 12x
20
e y = 4x3
r
13a 4.62 m
e Radius 2 cm, height 99 cm or radius 18.9 cm,
9a
b
c
64a + 16b + 4c = −2
d 0.77 m
(0, 9.86)
(6, 9.85)
9.8
R
9.7
(20, 9.71)
O
C
9.6
4
U
N
0
E
T(t) = – 0.00005(t – 6)3 + 9.85
R
T
9.9
height 11.5 cm
14a y = −0.164x3 + x2 − 1.872x + 2.1
b Length 3.464 units; width 6.000 units
EC
10a T(3) = 9.85, T(20) = 9.71
b
c Greatest volume of 4836.8 cm3; base radius 11.5 cm;
TE
D
125a + 50b + 20c = −8
b Proof required — check with your teacher.
PA
G
f
height 1.1 cm
4837 cm3
(0, 2.1), (1.25, 1), (2.5, 1.1), (4, 0.1)
d = 2.1
125a + 100b + 80c = −70.4
x
O(0, 0)
1 𝜋r3
V = 200𝜋r – –
2
0
D(3, 0)
FS
f
12a 3 x-intercepts; 2 turning points
Max volume when h is between 4 and 5 (estimates
will vary).
8
Height
≈ 4.6 m
!3
Proof required — check with your teacher.
Proof required — check with your teacher.
0 ≤ r ≤ 20
O
h
3
5 + !33
2
PR
O
0
e x = 4, x =
8
12
16
20
24 t
c T(28) = 9.32; unlikely, but not totally impossible.
Model is probably not a good predictor.
Topic 4 Cubic polynomials c04CubicPolynomials.indd 221
221
21/07/15 12:03 pm